Tanuja

Chapter 3 The Plane

Definition Of A Plane In Geometry With Examples And Proofs

Definition: A Plane is a surface such that if any two points are taken on it, the line joining them lies wholly on the surface.

Theorem.1 Every equation of the first degree in x, y, z represents a plane.

Proof. Let ax + bx + cz + d = 0, a₂ + b₂ + c₂ ≠ 0      …..(1)

be the first-degree equation in x, y, z.

If we have to show that (1) represents the equation to the plane, we prove that every point on the line joining any two points on (1) also lies on the locus (1).

Let p(x, y, z) and Q(x₂, y₂, z₂) be any two points on the locus (1).

Then we have ax₁ + bx₁ + cz₁ + d = 0 …..(2)

and ax₂ + by₂ + cz₂ + d = 0 …..(3)

Let R be any point on the line segment joining the points P and Q. Suppose R divides PQ in the ratio K : 1.

then \(\mathrm{R}=\left(\frac{\mathrm{K} x_2+x_1}{\mathrm{~K}+1}, \frac{\mathrm{K} y_2+y_1}{\mathrm{~K}+1}, \frac{\mathrm{K} z_2+z_1}{\mathrm{~K}+1}\right), \mathrm{K}+1 \neq 0\)

We have to show that R lies on the locus (1) for all values of K(≠ -1).

Substituting the coordinates of R in the LHS of (1), we get

⇒ \(\frac{a\left(\mathrm{~K} x_2+x_1\right)}{\mathrm{K}+1}+\frac{b\left(\mathrm{~K} y_2+y_1\right)}{\mathrm{K}+1}+\frac{c\left(\mathrm{~K} z_2+z_1\right)}{\mathrm{K}+1}+d\)

= a(Kx₂ + x₁) + b(Ky₂ + y₁) + c(Kz₂ + z₁) + d(K + 1)

= K(ax₂ + by₂ + cz₂ + d) + (ax₁ + by₁ + cz₁ + d) = K(0) + = 0 which shows that R lies on the locus (1).

Since R is an arbitrary point on the line joining P and Q, every point on PQ lies on (1)

∴ The equation ax + by + cz + d = 0, a₂ + b₂ + c₂ ≠ 0 always represents a plane.

Chapter 3 The Plane Converse Of The Above Theorem

Theorem.2 The equation to every plane is of the first degree in x, y, z.

Proof. Let π be the plane and O be the origin.

Case(1). Let O ∉ π, and let M be the foot of the perpendicular from O in π.

Let OM = p (>0). Let p(x, y, z) be any point in the plane.

Let [l, m, n] be Dc’s of the perpendicular OM.

p ≠ M. jOIN OP. OM = Projection of OP along OM

⇒ p = l(x-0)+m(y-0)+n(z-0) = lx+my+nz

Case.(2). Let O ∈ π, then p = 0.

p ∈ π <=> lx + my + nz = 0 <=> lx + my + nz = p where p = 0

Since any point P on π satisfies the equation lx + my + nz = p

it represents the equation of the plane.

The equation to π is lx + my + nz = p where p ≥ 0.

Hence the equation to the plane lx + my + nz = p is a first-degree equation in x, y, z.

Note.

1. If O ∈ π, equation to the plane is lx + my nz = 0.

2. lx + my + nz = p is called the normal form of the equation to the plane. Coefficients of x, y, z in the equation are l, m, n, and [l, m,n] are Dc’s of the normal OM to the plane, where p(≥ 0) is the distance of the origin to the plane.

3. An equation to the plane, in general, is taken as ax + by + cz + d = 0.

Theorems Related To Planes With Solved Problems Step-By-Step

Theorem: The Plane Transformation Of The Equation To The Plane Into The Normal Form

Let the equation to the plane be ax + by + cz + d = 0, a2 + b2 + c2 ≠ 0 …..(1)

We can take d ≥ 0 or d ≤ 0.

ax + by + cz + d = 0 <=> ax + by + cz = -d. Dividing by \(\sqrt{a^2+b^2+c^2}\), we get

⇒ \(\frac{-a}{\sqrt{a^2+b^2+c^2}}+\frac{-b}{\sqrt{a^2+b^2+c^2}}+\frac{-c}{\sqrt{a^2+b^2+c^2}}=\frac{d}{\sqrt{a^2+b^2+c^2}}\) …..(2)

or \(\frac{a}{\sqrt{a^2+b^2+c^2}} x+\frac{b}{\sqrt{a^2+b^2+c^2}} y+\frac{c}{\sqrt{a^2+b^2+c^2}} z=-\frac{d}{\sqrt{a^2+b^2+c^2}}\) …..(3)

This ((3) or (2)) is of the form lx + my + nz = p [p ≥ 0]

Where \([l=\pm \frac{a}{\sqrt{\sum a^2}}, m=\pm \frac{b}{\sqrt{\sum a^2}}, n=\pm \frac{c}{\sqrt{\sum a^2}}, p=\mp \frac{d}{\sqrt{\sum a^2}}]\)

∴ The normal form of the equation to the plane (1) is

⇒ \(\pm \frac{a x}{\sqrt{\sum a^2}} \pm \frac{b y}{\sqrt{\sum a^2}} \pm \frac{c z}{\sqrt{\sum a^2}}=\mp \frac{d}{\sqrt{\sum a^2}}(d \leq 0 \text { or } d \geq 0)\)

Note.

1. Direction ratios of a normal to the plane ax + by + cz + d = 0 are (a, b, c).

i.e., the coefficients of x, y, z in the equation.

2. Distance of the origin from the plane ax + by + cz + d = 0 is \(\frac{|d|}{\sqrt{\left(a^2+b^2+c^2\right)}}\)

3. First-degree equation in x, y, z without constant term <=> plane is passing through the origin.

4. abc ≠ 0 and \(\left(\frac{1}{a}\right) x+\left(\frac{1}{b}\right) y+\left(\frac{1}{c}\right) z+(-1)=0\). This equation represents a plane intersecting the x-axis in the point (a, 0, 0), intersecting the y-axis in the point (0, b, 0), and intersecting the z-axis in the point (0, 0, c).

Chapter 3 The Plane (1). Consider the equation lx + my = p (l≠0, m≠0) of a plane (π), d.cs. of a normal to it being l,m,0. Since 0, 0, 1 are dc.s. of the z-axis and l.0+m.0+0.1 = 0, the normal to π is perpendicular to the z-axis i.e., π is parallel to the z-axis.

Hence lx + my = p is the equation to a plane parallel to the z-axis.

Similarly, lx + my = p is the equation to a plane parallel to the y-axis and my + nz = p is the equation to a plane parallel to the x-axis.

(2) Consider the equation lx = p(l≠0) of a plane (π), d.cs. of a normal to it being l, 0, 0. Since o, 1, 0 are d.cs. of the y-axis i.e., π is parallel to the y-axis. Similarly, π is also parallel to the z-axis. Hence π is a plane parallel to the yz plane (x=0).

Similarly, my = p is a plane parallel to the zx plane (y=0) and nz = p is a plane parallel to the xy plane (z=0).

Orthogonal Projection Of Points And Lines On Planes Examples

Theorem.3 If the equation a₁x + b₁y + c₁z + d₁ = 0, a₂x + b₂y + c₂z + d₂ = 0 represents the same plane, then a₁:b₁:c₁:d₁ = a₂:b₂:c₂:d₂.

Proof. Given equations are  a₁x + b₁y + c₁z + d₁ = 0    …..(1)

a₂x + b₂y + c₂z + d₂ = 0 …..(2)

∴ (a_1, b_1, c₁), (a_2, b_2, c₂) are d.rs. of normals to the same plane.

Since the normals are either equal (coincident) or parallel,

we have a₁:a₂ = b₁:b₂ = c₁:c₂ = λ(say) (λ≠0) or (a_1, b_1, c₁)=λ(a_2, b_2, c₂)

Let (x₁, y₁, z₁) be any point in the plane represented by (1) and (2).

∴ d₁ = -(a₁x₁ + b₁y₁ + c₁z₁)

= -(a_1, b_1, c₁).(x₁, y₁, z₁) = -λ(a_2, b_2, c₂).(x₁, y₁, z₁)

= -λ(a₂x₁ + b₂y₁ + c₂z₁) = -λd₂

∴ a₁:a₂ = b₁:b₂ = c₁:c₂ = d₁:d₂

Chapter 3 Angles Between Two Planes

Definition. Angles between two planes are equal to the angles between their normals.

Angles between the planes a₁x + b₁y + c₁z = d₁  , a₂x + b₂y + c₂z = d₂

Let the equation to the planes be a₁x + b₁y + c₁z + d₁ = 0 …..(1)

a₂x + b₂y + c₂z + d₂ = 0 …..(2)

Dc’s of the normal to (1) =

⇒ \(m_1=\left(\frac{a_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}, \frac{b_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}, \frac{c_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}\right)\) and

Dc’s of the normal to (2) =

⇒ \(m_2=\left(\frac{a_2}{\sqrt{\left(a_2{ }^2+b_2{ }^2+c_2{ }^2\right)}}, \frac{b_2}{\sqrt{\left(a_2{ }^2+b_2{ }^2+c_2{ }^2\right)}}, \frac{c_2}{\sqrt{\left(a_2{ }^2+b_2{ }^2+c_2{ }^2\right)}}\right)\)

Let θ be one of the angles between the planes.

∴ θ = one of the angles between the normals m1, m2

=\(\cos ^{-1}\left(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)} \sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}\right)\)

The other angle between the planes is 180° – θ.

Cor.1. Condition of parallelism.

Planes are parallel ⇒ θ = 0° or 180° ⇒ \(\pm 1=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)} \sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}\)

⇒ \(\left(a_1^2+b_1^2+c_1^2\right)\left(a_2^2+b_2^2+c_2^2\right)=\left(a_1 a_2+b_1 b_2+c_1 c_2\right)^2\)

⇒ a₁²b₂² + a₁²c₂² + b₁²a₂² + b₁²c₂² + c₁²a₂² + c₁²b₂² – 2a₁a₂b₁b₂ – 2b₁b₂c₁c₂ – 2c₁c₂a₁a₂ = 0

⇒ \(\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2=0\)

⇒ \(a_1 b_2-a_2 b_1=b_1 c_2-b_2 c_1=c_1 a_2-c_2 a_1=0 \Rightarrow a_1: a_2=b_1: b_2=c_1: c_2\)

OR: Planes are parallel ⇒ their normals are parallel

⇒ d.rs of normals are proportional ⇒ a1:a2 = b1:b2 = c1:c2.

Cor. 2. Condition of perpendicularity.

Planes are perpendicular ⇒ θ = 90° ⇒ a1a2 + b1b2 + c1c2 = 0.

example. The plane x + 2y – 3z + 4 = 0 is perpendicular to the plane 2x + 5y + 4z + 1 = 0

since (1)(2) + (2)(5) + (-3)(4) = 0

OR: Planes are perpendicular ⇒ their normals are perpendicular

⇒ (a_1, b_1, c₁).(a_2, b_2, c₂) = 0 ⇒ a₁a₂ + b₁b₂ + c₁:c₂ = 0

Note.

1. The equation a₁x + b₁y + c₁z + d₁ = 0, a₁x + b₁y + c₁z + d₂ = 0 represent a pair of parallel planes.

2. A plane parallel to ax + by + cz + d = 0 is ax + by + cz + d = k, where k is an unknown real number.

example 1. The equation of the plane through the point (x₁, y₁, z₁) and parallel to the plane ax + by + cz + d = 0 is ax + by + cz = a₁x + b₁y + c₁z .

example 2. The normals to the plane as x – y + z – 1 = 0, 3x + 2y – z + 2 = 0 are perpendicular since (1)(3) + (-1)(2) + 1(-1) = 0.

Chapter 3 The Plane Determination Of A Plane Under Given Conditions

Consider the equation ax + by + cz + d = 0 of a plane. Since (a, b, c) ≠ (0, 0, 0), without loss of generally, we can take a≠0.

∴ Equation of the plane is \(x+\frac{b}{a} y+\frac{c}{a} z+\frac{d}{a}=0\)

∴ To know uniquely \(\frac{b}{a}, \frac{c}{a}, \frac{d}{a}\) we require three conditions.

For example, we can find the equation to a plane, if (1) three non-collinear points in the plane are given. (2) if two points in the plane and a plane perpendicular to the required plane (3) if one point in the plane and two planes perpendicular to the required plane are given.

Definition. If a plane π intersects the coordinate axes at (a, 0, 0), (0, b, 0), (0, 0, c) then a, b, c are respectively called the x-intercept, the y-intercept, the z-intercept of the plane π.

If the plane lx + my + nz = p intersects the x-axis at (a, 0, 0) then its x-intercept = a = \(\frac{p}{l}\).

Similarly its y-intercept = b = \(\frac{p}{m}\), its z-intercept = c = \(\frac{p}{n}\)

Note. If abc ≠ 0 and ax + by + cz + d = 0 …..(1) is a plane, then its x-intercept = \(-\frac{d}{a}\), (by putting y = 0, z = 0 in (1))

y-intercept = \(-\frac{d}{b}\), z-intercept = \(-\frac{d}{c}\).

example. The intercepts made by the plane x – 12y – 2z = 9 with the axes are \(\frac{9}{1},-\frac{9}{12},-\frac{9}{2}, \text { i.e., } 9,-\frac{3}{4},-\frac{9}{2}\).

Step-By-Step Solutions For Orthogonal Projection Problems On Planes

Theorem.4 Equation to the plane making intercepts a, b, c on the coordinate axes is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\).

Proof. Let π be the plane making intercepts,

a, b, c on the coordinate axes

Let A = (a, 0, 0), B = (0, b, 0), and C = (0, 0, c).

∴ abc ≠ 0.

Clearly, A, B, C are non-collinear. Let the equation to the plane π in the normal form be

lx + my + nz = p …..(1)

Let M be the foot of the perpendicular

from O to π and let [l, m, n] be the Dc’s of OM. Let OM = p.

p = OM = Projection of OA on OM = al.

Similarly p = bm, and p = cn. ∴ From(1)

equation to the plane π is \(\frac{p}{a} x+\frac{p}{b} y+\frac{p}{c} z=p \Rightarrow \frac{x}{a}+\frac{x}{b}+\frac{z}{c}=1\) (∵ p ≠ 0)

Note. Equation to the plane ABC is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\). This is called the intercept form of the equation to the plane and this plane does not pass through the origin.

OR: Let A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c). Let \(\mathrm{P}=\bar{r}=(x, y, z)\).

Now A, B, C are non-collinear. \(P \in \overleftrightarrow{A B C}(P \neq A \text { or } P=A)\)   (∵ abc ≠ 0)

<=> \(\overline{\mathrm{AP}}, \overline{\mathrm{AB}}, \overline{\mathrm{AC}}\) are coplanar or \(\overline{\mathrm{AP}}(=\overline{0}), \overline{\mathrm{AB}}, \overline{\mathrm{AC}}\) are three vectors.

<=> \(\bar{r}-\bar{a}(\neq \overline{0}), \bar{b}-\bar{a}, \bar{c}-\bar{a}\) are coplanar or \(\bar{r}-\bar{a}(=\overline{0}), \bar{b}-\bar{a}, \bar{c}-\bar{a}\) are three vectors.

<=> [(x-a, y, z),(-a, b, 0), (-a, 0, c)] = 0

<=> \(\left|\begin{array}{ccc}
x-a & y & z \\
-a & b & 0 \\
-a & 0 & c
\end{array}\right|=0 \Leftrightarrow(x-a) b c+y a c+z a b=0\)

<=> \(\frac{x}{a}-1+\frac{y}{b}+\frac{z}{c}=0 \)

<=> \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\).

∴ Equation to \(\overleftrightarrow{\mathrm{ABC}} \text { is } \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\).

Note. \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) is called the intercept form of the equation of the plane and this plane does not pass through the origin.

Properties Of Planes And Their Orthogonal Projections With Examples

Theorem. Equation to the plane determined by three non-collinear points A(x₁, y₁, z₁), B(x₂, y₂, z₂), C(x₃, y₃, z₃) is

⇒ \(\left|\begin{array}{llll}
x & y & z_1 & 1 \\
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1
\end{array}\right|=0\)

Proof. Let the equation of the required plane be ax + by + cz + d = 0 …..(1)

This passes through the given points if a₁x + b₁y + c₁z + d = 0 …..(2)

a₂x + b₂y + c₂z  + d = 0 …..(3) ax₃ + by₃ + cz₃ + d = 0 …..(4)

Eliminating a, b, c, d from the above equations (1), (2), (3), (4), we have

⇒ \(\left|\begin{array}{llll}
x & y & z & 1 \\
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1
\end{array}\right|=0\) …..(5)

This is the equation to the required plane.

OR :

Proof. Let \(\mathrm{A}=\bar{a}=\left(x_1, y_1, z_1\right), \quad \mathrm{B}=\bar{b}=\left(x_2, y_2, z_2\right), \mathrm{C}=\bar{c}=\left(x_3, y_3, z_3\right)\)

Let \(\bar{r}=(x, y, z) . \quad \bar{r}(\neq \bar{a}) \in \pi, \bar{r}(=\bar{a}) \in \pi\)

<=> \(\bar{r}-\bar{a}, \bar{b}-\bar{a}, \bar{c}-\bar{a}\) are coplanar or \(\bar{r}-\bar{a}(=\overline{0}), \bar{b}-\bar{a}, \bar{c}-\bar{a}\) are three vectors

<=> \( [\bar{r}-\bar{a}, \bar{b}-\bar{a}, \bar{c}-\bar{a}]=0\)

<=> \(\left[\left(x-x_1, y-y_1, z-z_1\right),\left(x_2-x_1, y_2-y_2, z_2-z_1\right),\left(x_3-x_1, y_3-y_1, z_3-z_1\right)\right]=0\)

<=> \(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
x_3-x_1 & y_3-y_1 & z_3-z_1
\end{array}\right|=0\) …..(1)

<=> \(\left|\begin{array}{cccc}
x-x_1 & y-y_1 & z-z_1 & 0 \\
x_1 & y_1 & z_1 & 1 \\
x_2-x_1 & y_2-y_1 & z_2-z_1 & 0 \\
x_3-x_1 & y_3-y_1 & z_3-z_1 & 0
\end{array}\right|=0\)

<=> \(\left|\begin{array}{cccc}
x & y & z & 1 \\
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1
\end{array}\right|=0 \mid \begin{aligned}
& \mathrm{R}_2+\mathrm{R}_1 \\
& \mathrm{R}_3+\mathrm{R}_1 \\
& \mathrm{R}_4+\mathrm{R}_1
\end{aligned}\)

Note. equation (1) may also be taken as the equation of the required plane.

Note. If the points (x₁, y₁, z₁), (x₂, y₂, z₂), (x₃, y₃, z₃), (x₄, y₄, z₄) are such that

⇒ \(\left|\begin{array}{llll}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
x_4 & y_4 & z_4 & 1
\end{array}\right|=0\), then the points are coplanar.

⇒ If \(\left|\begin{array}{llll}
x_1 & y_1 & z_1 & 1 \\
x_2 & y_2 & z_2 & 1 \\
x_3 & y_3 & z_3 & 1 \\
x_4 & y_4 & z_4 & 1
\end{array}\right| \neq 0\), then the points are non-coplanar.

Applications Of Orthogonal Projection On Planes In Mathematics

Theorem. Equation to the plane (π) through the point A(x₁, y₁, z₁) and perpendicular to the line (L) with d.rs. (a, b, c) is a(x-x₁)+b(y-y₁)+c(z-z₁) = 0.

Proof: Let p ∈ π and p = (x, y, z). A = (x₁, y₁, z₁)

and d.rs. of L are (a, b, c)

Now d.rs. of AP = (x – x1, y – y1, z – z1).

⇒ \(\overleftrightarrow{A P} \in \pi \Leftrightarrow \overleftrightarrow{A P} \perp L\)

<=> a(x-x1) + b(y-y1) + c(z-z1) = 0

∴ Equation to π is a(x – x1) + b(y – y1) + c(z – z1) = 0.

Chapter 3 The Plane Parametric Equation Of A Plane

Theorem. An equation to the plane passing through three non-collinear points A(a), B(b), C(c) is r = (1 – t – s) a + sb + tc, where s, t are any scalars (real numbers)

Proof: \(\bar{r} \in \overleftrightarrow{\mathrm{ABC}} \Rightarrow \bar{r}-\bar{a}, \bar{b}-\bar{a}, \bar{c}-\bar{a}\) are coplanar;

or \(\bar{r}-\bar{a}(\not 0), \bar{b}-\bar{a}, \bar{c}-\bar{a}\) are three vectors

<=> \(\bar{r}-\bar{a}=s(\bar{b}-\bar{a})+t(\bar{c}-\bar{a})\) where s, t are any scalars

<=> \(\bar{r}=(1-s-t) \bar{a}+s \bar{b}+t \bar{c}\) is the equation of the plane through \(\bar{a}, \bar{b}, \bar{c}\).

Note. Let \(\bar{r}=(x, y, z), \bar{a}=\left(x_1, y_1, z_1\right), \bar{b}=\left(x_2, y_2, z_2\right), \bar{c}=\left(x_3, y_3, z_3\right)\)

Then parametric equation to the plane \(\overleftrightarrow{\mathrm{ABC}}\) is

(x, y, z) = (1 – s – t)( x₁, y₁, z₁) + s(x₂, y₂, z₂) + t(x₃, y₃, z₃)

i.e. x = x₁ + s(x₂ – x₁) + t(x₃ – x₁),

y = y₁ + s(y₂ – y₁) + t(y₃ – y₁)

z = z₁ + s(z₂ – z₁) + t(z₃ – z₁)

Cor. Points \(\bar{a}, \bar{b}, \bar{c}, \bar{d}\) are coplanar

<=> \(\overline{\bar{d}}=(1-s-t) \bar{a}+s \bar{b}+t \bar{c}\)

<=> \((1-s-t) \bar{a}+s \bar{b}+t \bar{c}+(-1) \bar{d}=0\)

<=> \(\lambda_1 \bar{a}+\lambda_2 \bar{b}+\lambda_3 \bar{c}+\lambda_4 \bar{d}=0\),

λ₁ + λ₂ + λ₃ + λ₄= 1 – s – t + s + t – 1 = 0 and (λ₁ + λ₂ + λ₃ + λ₄) ≠ (0, 0, 0, 0)

Chapter 3 The Plane Two Sides Of A Plane

A Plane π divides the space into two half-spaces. Let a line \(\overleftrightarrow{\mathrm{AB}}\) intersect π in C. If A, B are in the same half-space, then (C; A, B) is negative, and A, B lie on the same side of C. If A, B are in different half spaces, then (C; A, B) is positive, and A, B lie on either side of C.

Theorem. If the line through A(x₁, y₁, z₁) and B(x₂, y₂, z₂) intersect the plane ax + by + cz + d = 0 in C, then (C; A, B) = -(a₁x + b₁y + c₁z + d):(a₂x + b₂y + c₂z + d).

Proof. Let A = (x₁, y₁, z₁) and B = (x₂, y₂, z₂) and C divide AB in the ratio λ1 : λ2 (λ1 + λ2 ≠ 0).

∴ (C; A, B) = λ₁ : λ₂ (λ₁ + λ₂ ≠ 0) and

⇒ \(\mathrm{C}=\left(\frac{\lambda_2 x_1+\lambda_1 x_2}{\lambda_1+\lambda_2}, \frac{\lambda_2 y_1+\lambda_1 y_2}{\lambda_1+\lambda_2}, \frac{\lambda_2 z_1+\lambda_1 z_2}{\lambda_1+\lambda_2}\right)\)

⇒ \(\mathrm{C} \in \pi \Leftrightarrow a\left(\frac{\lambda_2 x_1+\lambda_1 x_2}{\lambda_1+\lambda_2}\right)+b\left(\frac{\lambda_2 y_1+\lambda_1 y_2}{\lambda_1+\lambda_2}\right)+c\left(\frac{\lambda_2 z_1+\lambda_1 z_2}{\lambda_1+\lambda_2}\right)+d=0\)

<=> \(\lambda_2\left(a x_1+b y_1+c z_1+d\right)+\lambda_1\left(a x_2+b y_2+c z_2+d\right)=0\)

<=> \(\lambda_1\left(a x_2+b y_2+c z_2+d\right)=-\lambda_2\left(a x_1+b y_1+c z_1+d\right)\)

<=> \(\lambda_1: \lambda_2=-\left(a x_1+b y_1+c z_1+d\right):\left(a x_2+b y_2+c z_2+d\right)\)

<=> \((\mathrm{C} ; \mathrm{A}, \mathrm{B})=-\left(a x_1+b y_1+c z_1+d\right):\left(a x_2+b y_2+c z_2+d\right)\)

OR :
Proof: Let \(\mathrm{A}=\bar{a}=\left(x_1, y_1, z_1\right) \text { and } \mathrm{B}=\bar{b}=\left(x_2, y_2, z_2\right)\)

Let \(\mathrm{C}=\bar{c}\) divide AB in the ratio λ₁ : λ₂.

∴ (C; A, B) = λ₁ : λ₂ (λ₁ + λ₂  ≠ 0)

Then \(\bar{c}=\frac{\lambda_2 \bar{a}+\lambda_1 \bar{b}}{\lambda_1+\lambda_2}\)

Let the plane represented by ax + by + cz + d = 0 be \(\bar{r} \cdot \bar{m}=q\) where

⇒ \(\bar{r}=(x, y, z), \bar{m}=(a, b, c) \text { and } q=-d\)

∴ \(\mathrm{C} \in \pi \Leftrightarrow \bar{c} \cdot \bar{m}-q=0\)

<=> \(\left(\frac{\lambda_2 \bar{a}+\lambda_1 \bar{b}}{\lambda_1+\lambda_2}\right) \cdot \bar{m}-q=0\)

<=> \(\lambda_2(\bar{a} \cdot \bar{m})+\lambda_1(\bar{b} \cdot \bar{m})=q \lambda_1+q \lambda_2 \Leftrightarrow \lambda_1(\bar{b} \cdot \bar{m}-q)=-\lambda_2(\bar{a} \cdot \bar{m}-q)\)

<=> \(\lambda_1: \lambda_2=-(\bar{a} \cdot \bar{m}-q):(\bar{b} \cdot \bar{m}-q)\)

<=> \(\lambda_1: \lambda_2=-\left\{\left(x_1, y_1, z_1\right) \cdot(a, b, c)+d\right\}:\left\{\left(x_2, y_2, z_2\right) \cdot(a, b, c)+d\right\}\)

<=> \(\lambda_1: \lambda_2=-\left(a x_1+b y_1+c z_1+d\right):\left(a x_2+b y_2+c z_2+d\right)\)

<=> \((\mathrm{C} ; \mathrm{A}, \mathrm{B})=-\left(a x_1+b y_1+c z_1+d\right):\left(a x_2+b y_2+c z_2+d\right)\)

Note 1. A, B lie in the same half-space.

<=> a₁x + b₁y + c₁z + d, a₂x + b₂y + c₂z + d are of the same sign and A, B lie in the different half-spaces.

<=> a₁x + b₁y + c₁z + d, a₂x + b₂y + c₂z + d are of the different signs.

example

1. The points (1, 2, -5), (0, 4, -7) lie in the different half space (on the opposite sides) of the plane x + 2y + 2z – 9 = 0 since 2 + 2(3) + 2(5) – 9 > 0 and 0 + 2(4) + 2(-7) – 9 <0.

2. The points (1, 2, -5),(0, 4, -7) lie in the same half-space (on the same side) of the plane x + 2y + 2z – 9 = 0 since (1) + 2(2) + 2(-5) – 9 < 0 and 0 + 2(4) + 2(-7) – 9 < 0.

3. The points (1, -1, 3) and (3, 3, 3) lie on different sides of the plane x + 2y – 7z + 9 = 0 since 5(1) + 2(-) – 7(3) + 9 = -9 < 0 and 5(3) + 2(3) – 7(3) + 9 = 9 > 0.

Worked Examples Of Orthogonal Projection Onto Planes In Geometry

Chapter 3 The Plane Perpendicular Distance Of A Point From A Plane

Theorem. The distance of A(x₁, y₁, z₁) from the plane ax + by + cz + d = 0 i.e. length of the perpendicular from the point A(x₁, y₁, z₁) to the plane ax + by + cz + d = 0 is

⇒ \(\frac{\left|a x_1+b y_1+c z_1+d\right|}{\sqrt{\left(a^2+b^2+c^2\right)}}\).

Proof: Let π be the plane ax + by + cz + d = 0 …..(1)

Let A = (x₁, y₁, z₁) be the point (A ∉ π) from which the perpendicular drawn to the plane π meets it in C.

Let the normal form of π be lx + my + nz = p …..(2)

the equation to the plane parallel to (2) and passing through the point A be

lx + my + nz = p₁ …..(3) where lx₁ + my₁ + nz₁ = p₁          …..(4)

Let ODE be perpendicular to (2) and (3) as shown.

⇒ AC = p1 – p

⊥r distance of A to the plane π

= AC = OE – OD = lx₁ + my₁ + nz₁ – p

= \(+\frac{a}{\sqrt{\sum a^2}} x_1+\frac{b}{\sqrt{\sum a^2}} y_1+\frac{c}{\sqrt{\sum a^2}} z_1 \pm \frac{d}{\sqrt{\sum a^2}}\)

i.e., \(\pm \frac{\left(a x_1+b y_1+c z_1+d\right)}{\sqrt{a^2+b^2+c^2}} \text { or } \frac{\left|a x_1+b y_1+c z_1+d\right|}{\sqrt{a^2+b^2+c^2}}\)

OR : Proof: Let \(\mathrm{A}=\bar{a}=\left(x_1, y_1, z_1\right)\).

Let π be the plane ax + by + cz + d = 0.

Equation to π can be taken as \(\bar{r} \cdot \bar{m}=q \text { where } \bar{r}=(x, y, z)\).

⇒ \(\bar{m}=(a, b, c) \text { and } q=-d\).

Now, \(|\bar{m}|=\sqrt{\left(a^2+b^2+c^2\right)}\).

Let C be the foot of the perpendicular from A to π.

Let B(≠C) be \(\bar{b}\) in π.

∴ \(\bar{b} \cdot \bar{m}=q\) …..(1)

∴ \(\mathrm{AC}=\frac{|\overline{\mathrm{AB}} \cdot \overline{\mathrm{AC}}|}{|\overline{\mathrm{AC}}|}=\frac{|(\bar{b}-\bar{a}) \cdot \bar{m}|}{|\bar{m}|}=\frac{|\bar{b} \cdot \bar{m}-\bar{a} \cdot \bar{m}|}{|\bar{m}|}=\frac{|\bar{a} \cdot \bar{m}-\bar{b} \cdot \bar{m}|}{|\bar{m}|}=\frac{|\bar{a} \cdot \bar{m}-\bar{q}|}{|\bar{m}|}\)

= \(\frac{\left|\left(x_1, y_1, z_1\right) \cdot(a, b, c)-q\right|}{\sqrt{a^2+b^2+c^2}}=\frac{\left|a x_1+b y_1+c z_1+d\right|}{\sqrt{a^2+b^2+c^2}}\)

example. The distance of the points (2, 3, 4) and (1, 1, 4) from the plane 3x – 6y + 2z + 11 = 0

=\(\left|\frac{3(2)-6(3)+2(4)+11}{\sqrt{(9+36+4)}}\right|=1 \text { and }\left|\frac{3-6+8+11}{\sqrt{9+36+4}}\right|=\frac{16}{7}\)

Chapter 3 The Plane Distance Between Parallel Planes

Theorem. Distance between parallel planes ax + by + cz + d₁ = 0, ax + by + cz + d₂ = 0 is \(\frac{\left|d_1-d_2\right|}{\sqrt{a^2+b^2+c^2}}\), d₁ < 0, d₂ > 0.

Proof. The equations to the planes are

ax + by + cz + d₁ = 0 …..(1) ax + by + cz + d₂ = 0 …..(2)

the Dc’s of the normal to the planes are

⇒ \(\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)\)

Now p₁ and p₂ be the perpendicular distances to the planes from the origin

⇒ \(p_1=\frac{-d_1}{\sqrt{a^2+b^2+c^2}}, p_2=\frac{-d_2}{\sqrt{a^2+b^2+c^2}}\)

⇒ Distance between the parallel planes = \(\left|p_1-p_2\right|=\frac{\left|d_2-d_1\right|}{\sqrt{a^2+b^2+c^2}}\).

Note. Equation to the plane parallel to (1) and (2) and midway between (1) and (2)

(1) When the origin lies on the same side of both (1) and (2) is

i.e., \(a x+b y+c z=\frac{-\left(d_1+d_2\right)}{2}\)

(2) When origin lies in between (1) and (2) is \(ax+by+c z=\frac{\left|d_1-d_2\right|}{2}\)

example. The distance between the planes 2x – y + 3z = 6 and -6x + 3y – 9z = 5

= The distance between the planes 2x – y + 3z = 6 and \(2 x-y+3 z=-\frac{5}{3}\)

=\(\frac{\left|6-\left(-\frac{5}{3}\right)\right|}{\sqrt{(4+1+9)}}=\frac{23}{3 \sqrt{14}}\)

Chapter 3 The Plane Solved Problems

Example. 1. Find the point P equidistant from A(4, -3, 7) and B(2, -1, 1) and lying on y-axis. Hence find the equation to the plane through P and perpendicular to \(\overleftrightarrow{\mathrm{AB}}\).

Solution:

Given

A(4, -3, 7) and B(2, -1, 1)

Let P = (0, y, 0). Since PA = PB, PA² = PB²

⇒ (0 – 4)2 + (y + 3)2 + (0 – 7)2 = 4 + (y + 1)2 + 1 ⇒ 4y = -68 ⇒ y = -17

∴ P = (0, -17, 0) and d.rs of \(\overleftrightarrow{\mathrm{AB}}\) are 2, -2, 6.

∴ The equation to the plane through P, and perpendicular \(\overleftrightarrow{\mathrm{AB}}\) is

2(x – 0) – 2(y + 17) + 6(z – 0) = 0 i.e., 2x – 2y + 6z – 34 = 0,

i.e., x – y + 3z – 17 = 0

Geometric Interpretation Of Planes And Orthogonal Projections

Example.2. Show that the line joining the points (6, -4, 4), (0, 0, -4) intersects the line joining the points (-1, -2, -3), (1, 2, -5).

Solution.

Given

(6, -4, 4), (0, 0, -4)

And (-1, -2, -3), (1, 2, -5)

Let A = (6, -4, 4), B = (0, 0, 4), C = (-1, -2, -3), D = (1, 2, -5).

∴ \(\overline{\mathrm{AB}}=(-6,4,-8) \text { and } \overline{\mathrm{CD}}=(2,4,-2)\)

⇒ \(\overline{\mathrm{AB}}\) is neither perpendicular not parallel to \(\overline{\mathrm{CD}}\).

∴ If we prove that A, B, C, D are coplanar, then \(\overleftrightarrow{\mathrm{AB}}\) intersects \(\overleftrightarrow{\mathrm{CD}}\).

Now equation to the plane \(\overleftrightarrow{\mathrm{ABC}}\) is \(\left|\begin{array}{ccc}
x-6 & y+4 & z-4 \\
-6 & 4 & -8 \\
-7 & 2 & -7
\end{array}\right|=0\)

⇒ (-28 + 16)(x – 6) – (42 – 56)(y + 4) + (-12 + 28)(z – 4) = 0

⇒ 6(x – 6) – 7(y + 4) – 8(z – 4) = 0 ⇒ 6x – 7y – 8z – 32 = 0 …..(1)

Substituting D in the L.H.S of (1), we get 6 – 14 + 40 – 32 = 0 = R.H.S.

∴ \(D \in \overleftrightarrow{\mathrm{ABC}}\)

∴ \(\overleftrightarrow{\mathrm{AB}} \text { and } \overleftrightarrow{\mathrm{CD}}\) intersect.

Example.3. Obtain the equation to the plane containing (0, 4, 3) and the line through the points (-1, -5, -3), (-2, -2, 1). Hence show that (0, 4, 3), (-1, -5, -3), (-2, -2, 1) and (1, 1, -1) are coplanar.

Solution.

Given

(0, 4, 3) (-1, -5, -3), (-2, -2, 1)

Let π be the required plane. Let a, b, c be d.rs. of a normal to it.

Let A = (0, 4, 3), B = (-1, -5, -3), C = (-2, -2, 1)

D.rs. of \(\overleftrightarrow{\mathrm{AB}}\) are -1, -9, -6 and d.rs. of \(\overleftrightarrow{\mathrm{AC}}\) are -2, -6, -2.

Since \(\overleftrightarrow{\mathrm{AB}}, \overleftrightarrow{\mathrm{AC}}\) are in π,

⇒ \(\left.\begin{array}{r}
-a-9 b-6 c=0 \\
-2 a-6 b-2 c=0
\end{array}\right\} \text {. Solving, } \frac{a}{-18}=\frac{b}{10}=\frac{c}{-12} \text { i.e., } \quad \frac{a}{9}=\frac{b}{-5}=\frac{c}{6}\)

∴ Equation to π is 9(x – 0) – 5(y – 4) + 6(z – 3) = 0 i.e., 9x – 5y + 6z + 2 = 0 …..(1)

Clearly (1, 1, -1) lies on (1). Hence the points are coplanar.

∴ The equation to the plane containing the points is (1).

Example.4. Find the equation of the plane through (4, 4, 0) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 3z – 8 = 0.

Solution.

Given

(4, 4, 0)

x + 2y + 2z = 5 and 3x + 3y + 3z – 8 = 0

Let π be the required plane. Let a, b, c be d.rs of normal to π.

Since π passes through (4, 4, 0) equation to π is a(x – 4) + b(y – 4) + c(z – 0) = 0

But π is perpendicular to x + 2y + 2z = 5 and 3x + 3y + 2z – 8 = 0

∴ \(\left.\begin{array}{r}
a+2 b+2 c=0 \\
3 a+3 b+2 c=0
\end{array}\right\}\).

∴ \(\frac{a}{-2}=\frac{b}{4}=\frac{c}{-3}\).

∴ Equation to π is -2(x – 4) + 4(y – 4)-3z = 0 [using(1)]

i.e., 2x – 4y + 3z + 8 = 0.

Solved Problems On Equations Of Planes And Orthogonal Projection

Example.5. Find the equation of the plane passing through (1, 0, -2) and perpendicular to the planes 2x + y – 2 = z; x – y – z = 3

Solution.

Given

(1, 0, -2)

2x + y – 2 = z; x – y – z = 3

Let π be the required plane. Let a, b, c be the drs. of the above plane.

The equation of the plane passing through (1, 0, -2) and having a, b, c as drs. is a(x-1) + b(y-0) + c(z+2) = 0 ⇒ a(x-1) + by + c(z+2) = 0 …..(1)

But the π plane is perpendicular to the planes 2x + y – z = 2 and x – y – z = 3.

∴ 2a + b – c = 0 …..(2), a – b – c = o …..(3)

Solving (2) and (3) \(\frac{a}{2}=\frac{b}{-1}=\frac{c}{3}\).

Equation of the π plane is 2(x-1)-y+3(z+2)=0 i.e., 2x – y + 3z + 4 = 0.

Example.6. Find the angles between the planes 2x – 3y – 6z = 6 and 6x + 3y – 2z = 18.

Solution.

Given Planes

2x – 3y – 6z = 6 and 6x + 3y – 2z = 18.

Let θ be one of the angles between the given planes.

∴ \(\theta=\cos ^{-1} \frac{2(6)-3(3)-6(-2)}{\sqrt{(4+9+36)} \sqrt{(36+9+4)}}=\cos ^{-1}\left(\frac{15}{49}\right)\)

The other angle between the planes is 180° – θ i.e., \(180^{\circ}-\cos ^{-1}\left(\frac{15}{49}\right)\)

Example.7. Find the locus of the point whose distance from the origin is three times its distance from the plane 2x – y + 2z = 3.

Solution.

Given

2x – y + 2z = 3

Let O be the origin and P be the point (x₁, y₁, z₁) such that OP is equal to 3 times its distance from the plane 2x – y + 2z = 3

∴ \(\mathrm{OP}^2=9 \cdot \frac{\left(2 x_1-y_1+2 z_1-3\right)^2}{4+1+4}\)

⇒ x₁² + y₁² + z₁² = 4x₁² + y₁² + 4z₁² + 9 – 4x₁y₁ – 4y₁z₁ + 8x₁z₁ – 12x₁ + 6y₁ – 12z₁

⇒ 3x₁² + 3z₁² – 4x₁y₁ – 4y₁z₁ + 8x₁z₁ – 12x₁ + 6y₁ – 12z₁ + 9 = 0

∴ Locus of P is 3x² + 3z² – 4xy – 4yz + 8xy – 12x + 6y – 12z + 9 = 0

Chapter 3 The Plane Systems Of Planes

Consider the equation ax + by + cz + d = 0, (a, b, c) ≠ (0, 0, 0),\(\lambda_1=\frac{b}{a}, \lambda_2=\frac{c}{a}, \lambda_3=\frac{d}{a}\) of a plane.

When three conditions satisfying the equation are given, λ₁, λ₂, λ₃ can be uniquely determined and hence a plane can be uniquely determined.

When two conditions satisfying the equation are given, one of λ₁, λ₂, λ₃ say, λ₁ cannot be found uniquely and λ₁ is called a parameter. Since λ₁ can be assigned any real vaule, an infinite number of planes arise, and these planes are called a system of planes.

When one condition satisfying the equation is given we have two parameters, say, λ₁, λ₂ giving rise to a system of planes for different values of λ₁, λ₂.

We give below a few systems of planes involving one or two parameters.

(1) The equation ax + by + cz + λ = 0 represents the system of planes parallel to a given plane ax + by + cz + d = 0, λ being the parameter.

(2) The equation ax + by + cz + λ = 0 represents the system of planes perpendicular to lines with d.rs. a, b, c; λ being the parameter.

(3) The equation a(x – x₁)+b(y – y₁)+c(z – z₁) = 0, (a, b, c) ≠ (0, 0, 0)

i.e., (x – x₁) + λ₁ (y – y₁) + λ₂ (z – z₁)= 0 where a ≠ 0(say), λ₁ = (b/a), λ₂= (c/a) represents the system of planes passing through the point (x₁, y₁, z₁); λ₁, λ₂ being the parameters.

(4) The equation λ₁(a₁x + b₁y + c₁z + d₁ ) + λ₂ (a₂x + b₂y + c₂z + d₂) = 0 represents the system of planes through the line of intersection of the planes a₁x + b₁y + c₁z + d₁ = 0, a₂x + b₂y + c₂z + d₂ = 0; λ₁, λ₂ being parameters and (λ₁, λ₂) ≠ (0, 0).

The truth of the statement can be seen from the theorem proved in the ensuing article.

Theorem. π₁ ≡ a₁x + b₁y + c₁z + d₁ = 0, π₂≡ a₂x + b₂y + c₂z + d₂ = 0 represent two intersecting planes.

(A) If (λ₁, λ₂) ≠ (0, 0), then λ₁π₁ + λ₂π₂ = 0 represents a plane passing through the line L of the intersection of π₁ and π₂.

(B) Any plane passing through the line L of the intersection of π₁ and π₂ is given by λ₁π₁ + λ₂π₂ = 0, (λ₁, λ₂) ≠ (0, 0)

Proof: (A) Let S ≡ λ₁π₁ + λ₂π₂ = 0, (λ₁, λ₂) ≠ (0, 0)

S is a first-degree equation in x, y, z and hence represents a plane.

Now λ₁ = 0 ⇒ S = π₂, λ₂ = 0 ⇒ S = π₁. Let P(x₁, y₁, z₁) ∈ L

∴ a₁x₁ + b₁y₁ + c₁z₁ + d₁ = 0 …..(1) a₂x₁ + b₂y₁ + c₂z₁ + d₂ = 0 …..(2)

Also from (1) and (2), P ∈ S, when (λ₁, λ₂) ≠ (0, 0)

∴ S represents a plane through the line L of the intersection of π₁ and π₂.

If  λ₁ ≠ 0, λ₂ ≠ 0, for different values of λ₁, λ₂; S represents any plane through the line L of intersection π₁ and π₂ and different from π₁ and π₂.

(B) Let P(x₁, y₁, z₁), Q(x₂, y₂, z₂) be different points on L such that x₁ ≠ x₂ (say).

Let S ≡ αx + βy + γz + δ = 0 be a plane through L and hence

αx₁ + βy₁ + γz₁ + δ = 0 …..(3)  αx₂ + βy₂ + γz₂ + δ = 0 …..(4)

Let l, m, n be d.rs. of L, the line of intersection of the planes π₁ and π₂.

∴ \(\left.\begin{array}{c}
a_1 l+b_1 m+c_1 n=0 \\
a_2 l+b_2 m+c_2 n=0
\end{array}\right\} \frac{l}{b_1 c_2-b_2 c_1}=\frac{m}{c_1 a_2-c_2 a_1}=\frac{n}{a_1 b_2-a_2 b_1}\)

Since (b₁c₂ – b₂c₁, c₁a₂ – c₂a₁, a₁b₂ – a₂b₁) ≠ (0, 0, 0) without loss of generality we can take b₁c₂ – b₂c₁ ≠ 0.

For λ₁, λ₂ and (λ₁, λ₂) ≠ (0,0) there exist equations λ₁b₁ + λ₂b₂ = β, λ₁c₁ + λ₂c₂ = γ

such that they have a unique solution λ₁ and  λ₂.

∴ αx + βy + γz + δ ≡ αx + (λ₁b₁ + λ₂b₂)y (λ₁c₁ + λ₂c₂)z + δ

≡ λ₁(a₁x + b₁y + c₁z + d₁) + λ₂(a₂x + b₂y + c₂z + d₂) + αx – λ₁a₁x – λ₂a₂x – λ₁d₁ – λ₂d₂ + δ

≡ λ₁(a₁x + b₁y + c₁z + d₁) + λ₂(a₂x + b₂y + c₂z + d₂) + (α – λ₁a₁– λ₂a₂) x + (δ – λ₁d₁ – λ₂d₂)

≡ λ₁(a₁x + b₁y + c₁z + d₁) + λ₂(a₂x + b₂y + c₂z + d₂) + λ₃x + λ₄    …..(5)

Where λ3 = α – λ₁a₁– λ₂a₂, λ₄ = δ – λ₁d₁ – λ₂d₂

∴ P ∈ L ⇒ P ∈ S ⇒ αx₁ + βy₁ + γz₁ + δ = 0 ⇒ λ₃x₁ + λ₄= 0 …..(6)

using (3) and (5), and Q ∈ L ⇒ Q ∈ S ⇒ λ₃x₂ + λ₄ = 0 …..(7)

using (4) and (5),

∴ (6) – (7) ⇒ λ₃(x₁ – x₂) = 0 ⇒ λ₃ = 0 (∵ x₁ ≠ x₂)

∴ From (6), λ₄ = 0.

∴ S ≡ λ₁π₁ + λ₂π₂ = 0 is the plane passing through the line of intersection π₁ and π₂.

Note. Let λ₁ ≠ 0 (say). Now equation to the plane (distinct from π₁, π₂) passing through the line of intersection of planes π₁ and π₂ can be taken as \(\pi_1+\left(\frac{\lambda_2}{\lambda_1}\right) \pi_2=0\) i.e. π₁ + λπ₂ = 0 where \(\lambda=\frac{\lambda_2}{\lambda_1}\). This form of equation might be taken while doing problems.

Chapter 3 The Plane Planes Bisecting The Angles Between Two Planes.

Theorem.  π₁ ≡ a₁x + b₁y + c₁z + d₁= 0, π₂ ≡ a₂x + b₂y + c₂z + d₂ = 0 and d_1d2 > 0. Equation to the plane bisecting the angle containing the origin between the planes

⇒ \(\pi_1, \pi_2 \text { is } \frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=+\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}\)

and to the plane bisecting the other angle between the planes

⇒ \(\pi_1, \pi_2 \text { is } \frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=-\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}\)

Proof: Equations to π₁, π₂ are

a₁x + b₁y + c₁z + d₁ = 0 …..(1)   a₂x + b₂y + c₂z + d₂ = 0 …..(2) and d₁ d₂>0

we know that if p(x, y, z) is any point on one of the planes bisecting the angle between π₁, π₂ then the perpendicular distances of P from π₁, π₂ are equal (in magnitude) so that

⇒ \(\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=\pm \frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}\) are the equations to the bisecting planes.

∴ The equation to the plane bisecting the angle containing the origin is

⇒ \(\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=+\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}} \text { if } d_1>0, d_2>0\) …..(7)

This plane bisects the angle containing the origin also bisects the vertically opposite angle.

∴ The equation to the plane bisecting the other angle and its vertically opposite angle is

⇒ \(\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{a_1^2+b_1^2+c_1^2}}=-\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}} \text { if } d_1>0, d_2>0\) …..(8)

Note 1. In bisecting planes (7) and (8), one bisects the acute and the other bisects the obtuse angle between the given planes π₁, π_2.

The bisecting plane of the acute angle makes with either of the planes π₁, π₂ an angle less than 45° and the bisecting plane of the obtuse angle makes with either of the planes π₁, π₂ an angle greater than 45° (of course < 90°). This gives a test for determining which angle (acute or obtuse) each bisecting plane bisects.

2. Even if d₁ < 0, d₂ < 0, the theorem holds.

But if d₁ > 0, d₂ < 0 or d₁ < 0, d₂ > 0, equation to the plane bisecting the angle containing the origin is \(\frac{a_1 x+b_1 y+c_1 z+d_2}{\sqrt{a_1^2+b_1^2+c_1^2}}=-\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{a_2^2+b_2^2+c_2^2}}\) and the other equation gives the other bisecting plane.

3. l₁x + m₁y + n₁z = q₁, l₂x + m₂y + n₂z = q₂ are two intersecting planes such that (l₁, m₁, n₁), (l₂, m₂, n₂) are unit points and q₁q₂ > 0 (q_{1 ,} q₂ are of the same sign).

∴ The equation to the plane bisecting the angle containing the origin is (l₁ – l₂)x + (m₁ – m₂)y + (n₁ – n₂)z = q₁ – q₂ and the equation to the plane bisecting the other angle is (l₁ + l₂)x + (m₁ + m₂)y + (n₁ + n₂)z = q₁ + q₂

Chapter 3 The Plane Solved Problems

Example. 1. Find the equation to the plane through the point (x₁, y₁, z₁) and parallel to the plane ax + by + cz + d = 0

Solution.

Given point (x₁, y₁, z₁) and plane ax + by + cz + d = 0

Let ax + by + cz + λ = 0 …..(1)

be the plane parallel to ax + by + cz + d = 0 …..(2)

for all values of λ.

If (1) passes through (x₁, y₁, z₁) then ax₁ + by₁ + cz₁ + λ = 0 i.e., λ = – ax₁ – by₁ – cz₁

∴ Required plane is ax + by + cz – ax₁ – by₁ – cz₁ = 0

i.e., a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

Example. 2. Find the equations of the planes through the intersection of the planes x + 3y + 6 = 0 and 3x – y – 4z = 0 such that the perpendicular distances of each from the origin are unity.

Solution.

Given planes x + 3y + 6 = 0 and 3x – y – 4z = 0

Let the plane passing through the intersection of the planes x + 3y + 6 = 0, 3x – y – 4z = 0 be (x + 3y + 6) + λ(3x – y – 4z) = 0

⇒ (1 + 3λ)x + (3 – λ)y – 4λz + 6 = 0 …..(1)

Perpendicular distance of origin from (1) = 1

⇒ \(\frac{6}{\sqrt{(1+3 \lambda)^2+(3-\lambda)^2+16 \lambda^2}}=1 \Rightarrow 26 \lambda^2=26 \Rightarrow \lambda=\pm 1\)

∴ Required planes are 4x + 2y – 4z + 6 = 0, -2x + 4y + 4z + 6 = 0

i.e., 2x + y – 2z + 3 = 0, x – 2y – 2z – 3 = 0

Example. 3. Find the equation to the plane through the intersection of the planes x + 2y + 3z + 4 = 0 and 4x + 3y + 3z + 1 = 0 and perpendicular to the plane x + y + z + 9 = 0

Solution.

Given planes x + 2y + 3z + 4 = 0 and 4x + 3y + 3z + 1 = 0 and perpendicular to the plane x + y + z + 9 = 0

Let the plane through the intersection of the planes

x + 2y + 3z + 4 = 0, 4x + 3y + 3z + 1 = 0 be (x + 2y + 3z + 4) + λ(4x + 3y + 3z + 1) = 0

⇒ (1 + 4λ)x + (2 + 3λ)y + (3 + 3λ)z + (4 + λ) = 0 …..(1)

If (1) is perpendicular to x + y + z + 9 = 0, then (1 + 4λ).1 + (2 + 3λ).1 + (3 + 3λ).1 = 0

i.e., 10λ = -6 i.e., λ = -3/5.

∴ Required plane is 7x – y – 6z – 17 = 0

Example. 4. Find the equation to the plane through the line of intersection of x – y + 3z + 5 = 0 and 2x + y – 2z + 6 = 0 and passing through (-3, 1,1)

Solution.

Given Planes x – y + 3z + 5 = 0 and 2x + y – 2z + 6 = 0 and passing through (-3, 1,1)

Let the equation to the plane through the intersection of the planes x – y + 3z + 5 = 0, be (x – y + 3z + 5) + λ(2x + y – 2z + 6) = 0 …..(1), for and λ.

Let (1) pass through the point (-3, 1,1).

∴ -3 – 1 + 3 + 5 + λ(-6 + 1 -2 + 6) = 0 i.e., -λ + 4 = 0 i.e., λ = 4

∴ Equation to the required plane is 9x + 3y – 5z + 29 = 0

Example. 5. Find the equation to the plane through (2, -3, ) and is normal to the line joining (3, 4, -1) and (2, -1, 5).

Solution.

Given

(2, -3, ) (3, 4, -1) and (2, -1, 5)

Let the plane through (2, -3, 1) and perpendicular to the join of p(3, 4, -1) and Q(2, -1, 5) be a(x – 2) + b(y + 3) + c(z – 1) = 0.

Since d.rs of \(\overleftrightarrow{\mathrm{PQ}}\) are (3 – 2, 4 + 1, -1 -5) i.e., (1, 5, -6)

We have \(\frac{a}{1}=\frac{b}{5}=\frac{c}{-6}=\lambda, \text { say }\).

∴ Required plane is λ(x – 2) + 5λ(y + 3) – 6λ(z – 1) = 0

i.e., x – 2 + 5(y + 3) – 6(z – 1) = 0 i.e., x + 5y – 6z = -19

Example. 6. A variable plane passes through a fixed point (a, b, c). It meets the axes of reference in A, B, and C. Show that the locus of the point of intersection of the planes through A, B, C, and parallel to the coordinate planes is ax^-1 + by^-1 + cz^-1 = 1

Solution.

Given

A variable plane passes through a fixed point (a, b, c). It meets the axes of reference in A, B, and C.

Let the variable plane meeting the coordinates axes in A, B,C be \(\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=1\)     …..(1)

∴ A = (α, 0, 0), B = (0, β, 0), C = (0, 0, γ)

Also (1) passes through the fixed point (a, b, c)

∴ \(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}=1\)

But equations to the planes through A, B, C and parallel to the coordinate planes are x = α, y = β, z = γ.

Clearly they intersect at P = (α, β, γ)

∴ Locus of P from (2) is \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1 \text { i.e., } ax^{-1}+b y^{-1}+c z^{-1}=1\)

Example. 7. Find the bisecting plane of the acute angle between the planes 3x – 2y – 6z + 2 = 0, -2x + y – 2z – 2 = 0

Solution.

Equations to the given planes are taken as 3x – 2y – 6z + 2 = 0 …..(1)

2x – y + 2x + 2 = 0 …..(2)

(constant terms are taken as +ve)

∴ Equations to the bisecting planes between the given planes are

⇒ \(\frac{3 x-2 y+6 z+2}{\sqrt{(9+4+36)}}=\pm \frac{2 x-y+2 z+2}{\sqrt{(4+1+4)}}\)

i.e., 5x – y – 4z + 8 = 0 …..(3)

23x – 13y + 32z + 20 = 0 …..(4)

Let θ be the acute angle between (2) and (3)

∴ \(\cos \theta=\left|\frac{10+1-8}{\sqrt{9} \cdot \sqrt{(25+1+16)}}\right|=\frac{1}{(\sqrt{42})}\)

∴ \(\tan \theta=\sqrt{(41)}>1\)

Hence 2θ, the angle between the planes

(1) and (2) are greater than 90° i.e., obtuse.

∴ (3) is the equation to the plane bisecting the obtuse angle between (1) and (2).

∴ (4) is the equation to the plane bisecting the acute angle between (1) and (2).

Note. 5x – y – 4z + 8 = 0 is the plane bisecting the angle containing the origin between (1) and (2).

Chapter 3 The Plane Joint Equation Of A Pair Of Planes

Consider the pair of planes π₁, π₂ whose respective equations are

l₁x + m₁y + n₁z + d₁ = 0 …..(1) l₂x + m₂y + n₂z + d₂ = 0 …..(2)

Consider the equation (l₁x + m₁y + n₁z + d₁)(l₂x + m₂y + n₂z + d₂) = 0 …..(3)

Let P = (x₁, y₁, z₁).

P ∈ π₁ ⇒ l₁x₁ + m₁y₁ + n₁z₁ + d₁ = θ ⇒ (l₁x₁ + m₁y₁ + n₁z₁ + d₁)(l₂x₁ + m₂y₁ + n₂z₁ + d₂) = 0

P ∈ π₂ ⇒ l₂x₁ + m₂y₁ + n₂z₁ + d₂ = θ ⇒ (l₁x₁ + m₁y₁ + n₁z₁ + d₁)(l₂x₁ + m₂y₁ + n₂z₁ + d₂) = 0

P lies on (3) ⇒ (l₁x₁ + m₁y₁ + n₁z₁ + d₁)(l₂x₁ + m₂y₁ + n₂z₁ + d₂) = 0

⇒ l₁x₁ + m₁y₁ + n₁z₁ + d₁ = 0 or ⇒ l₂x₁ + m₂y₁ + n₂z₁ + d₂ = 0 ⇒ P ∈ π₁ or ⇒ P ∈ π₂

∴ We have P ∈ π₁ or P ∈ π₂ : P lies on (3)

i.e., an equation is satisfied if and only if a point lies on the one plane or the other plane or both.

∴ (3) represents the joint or combined equation to the plane π₁ and π₂.

Note 1. a₁x + b₁y + c₁z + d₁ = 0, a₂x + b₂y + c₂z + d₂ = 0 are two intersecting planes.

The combined equation to the pair of planes bisecting the angles between them is

⇒ \(\left[\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}-\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}\right]\)

⇒ \(\left[\frac{a_1 x+b_1 y+c_1 z+d_1}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)}}+\frac{a_2 x+b_2 y+c_2 z+d_2}{\sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}\right]=0\)

i.e., \(\frac{\left(a_1 x+b_1 y+c_1 z+d_1\right)^2}{a_1^2+b_1^2+c_1^2}-\frac{\left(a_2 x+b_2 y+c_2 z+d_2\right)^2}{a_2^2+b_2^2+c_2^2}=0\)

Theorem. S ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 represents a pair of planes π₁, π₂.

Then H ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 represents a pair of planes through the origin and parallel to the planes π₁, π₂.

Proof. Let the planes π₁, π₂ represented by S = 0 be respectively l₁x + m₁y + n₁z + d₁ = 0, l₂x + m₂y + n₂z + d₂ = 0

∴ ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d

≡ (l₁x + m₁y + n₁z + d₁)(l₂x + m₂y + n₂z + d₂)

⇒ l₁l₂ = a,m₁m₂ = b,n₁n₂ = c, l₁m₂ + l₂m₁ = 2h, m₁n₂ + m₂n₁ = 2f, n₁l₂ + n₂l₁ = 2g.

Joint equation of the planes passing through the origin and parallel to π₁, π₂ is

(l₁x + m₁y + n₁z)(l₂x + m₂y + n₂z) = 0

i.e., l₁l₂x² + m₁m₂y² + n₁n₂z² + (l₁m₂ + l₂m₁)xy + (m₁n₂ + m₂n₁)yz + (n₁l₂ + n₂l₁)zx = 0

i.e.,  ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 i.e., H = 0.

Definition. If H ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 and D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\), then D is called the determinant of H.

Theorem. H ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 represents the equation of a pair of planes or a plane if D = 0, f² ≥ bc, g² ≥ac, h² ≥ ab.

Proof. H = 0 represents the equation to a pair of planes or a plane ⇒ H can be expressed as a product of two linear factors in x, y, z.

Let the factors be l₁x + m₁y + n₁z + d₁, l₂x + m₂y + n₂z + d₂

where (l₁, m₁, n₁) ≠ (0, 0, 0)

∴ ax² + by² + cz² + 2fyz + 2gzx + 2hxy ≡ (l₁x + m₁y + n₁z + d₁)(l₂x + m₂y + n₂z + d₂)

⇒ l₁l₂ = a,m₁m₂ = b,n₁n₂ = c, l₁m₂ + l₂m₁ = 2h, m₁n₂ + m₂n₁ = 2f, n₁l₂ + n₂l₁ = 2g.

l₁d₂ + l₂d₁ = 0, m₁d₂ + m₂d₁ = 0, n₁d₂ + n₂d₁ = 0, d₁d₂ = 0

Now d₁d₂ = 0 ⇒ d₁ = 0 or d₂ = 0.

d₂ = 0 ⇒ l₂d₁ = 0, m₂d₁ = 0, n₂d₁ = 0

⇒ d₁ = 0 (∵ at least one of l₂, m₂, n₂ is not equal to zero)

Similarly d₁ = 0 ⇒ d₂ = 0. ∴ d₁ = d₂ = 0

We know that \(\left[\begin{array}{ccc}
l_1 & l_2 & 0 \\
m_1 & m_2 & 0 \\
n_1 & n_2 & 0
\end{array}\right]\left[\begin{array}{ccc}
l_2 & m_2 & n_2 \\
l_1 & m_1 & n_1 \\
0 & 0 & 0
\end{array}\right]\)

= \(\left[\begin{array}{ccc}
2 l_1 l_2 & l_1 m_2+l_2 m_1 & l_1 n_2+l_2 n_1 \\
l_2 m_1+l_1 m_2 & 2 m_1 m_2 & m_1 n_2+m_2 n_1 \\
n_1 l_2+l_1 n_2 & n_1 m_2+n_2 m_1 & 2 n_1 n_2
\end{array}\right]=\left[\begin{array}{ccc}
2 a & 2 h & 2 g \\
2 h & 2 b & 2 f \\
2 g & 2 f & 2 c
\end{array}\right]\)

∴ \(\operatorname{det}\left\{\left[\begin{array}{ccc}
l_1 & l_2 & 0 \\
m_1 & m_2 & 0 \\
n_1 & n_2 & 0
\end{array}\right]\left[\begin{array}{ccc}
l_2 & m_2 & n_2 \\
l_1 & m_1 & n_1 \\
0 & 0 & 0
\end{array}\right]\right\}=\operatorname{det}\left[\begin{array}{ccc}
2 a & 2 h & 2 g \\
2 h & 2 b & 2 f \\
2 g & 2 f & 2 c
\end{array}\right]\)

⇒ \(\left|\begin{array}{ccc}
l_1 & l_2 & 0 \\
m_1 & m_2 & 0 \\
n_1 & n_2 & 0
\end{array}\right|\left|\begin{array}{ccc}
l_1 & m_2 & n_2 \\
l_1 & m_1 & n_1 \\
0 & 0 & 0
\end{array}\right|=8\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & e
\end{array}\right| \Rightarrow 0 \times 0=8 \mathrm{D} \Rightarrow \mathrm{D}=0\).

Also, 4f² – 4bc = (m₁n₂ + m₂n₁)² – 4m₁n₂m₂n₁ = (m₁n₂ – m₂n₁)² ≥ 0 ⇒ f² ≥ bc.

Similarly, we can prove that g² ≥ ac, h² ≥ ab.

We give below two theorems, for which proofs may be supplied by the readers if needed.

1. H = 0  represents a pair of planes if (1) D = 0 and (2) at least one of h² – ab, f² – bc, g² – ac is + ve and the remaining two are non-negative.

2. H = 0 represents a plane if D = 0, h² – ab, f² – bc, g² – ac. In this case, H takes the form (a₁x + b₁y + c₁z)².

Theorem. If θ(≤ π/2) is the angle between the pair of planes H = 0, then \(\cos \theta=\left|\frac{a+b+c}{\sqrt{\left\{(a+b+c)^2+4\left(f^2+g^2+h^2-a b-b c-c a\right)\right\}}}\right|\)

Proof. Let the pair of planes represented by H = 0 be

l₁x + m₁y + n₁z = 0, l₂x + m₂y + n₂z = 0.

∴ ax² + by² + cz² + 2fyz + 2gzx + 2hxy ≡ (l₁x + m₁y + n₁z)(l₂x + m₂y + n₂z)

⇒ l₁l₂ = a, m₁m₂ = b, n₁n₂ = c

l₁m₂ + l₂m₁ – 2h, m₁n₂ + m₂n₁ = 2f, n₁l₂ + n₂l₁ = 2g

Since θ(≤ (π/2)) is the angle between the planes,

⇒ \(\cos \theta=\mid \frac{l_1 l_2+m_1}{\sqrt{\left(l_1^2+m_1^2+n_1^2\right)}}\frac{m_2+n_1 n_2}{\sqrt{\left(l_2^2+m_2^2+n_2^2\right)}}|\)

=\(\left|\frac{a+b+c}{\sqrt{\left\{\left(a^2+b^2+c^2+4 h^2-2 a b+4 f^2-2 b c+4 g^2-2 a c\right)\right\}}}\right|\)

=\(\left|\frac{a+b+c}{\sqrt{\left\{(a+b+c)^2+4\left(f^2+g^2+h^2-b c-c a-a b\right)\right\}}}\right|\)

Cor. 1. Planes are perpendicular <=> θ = 90°

<=> cosθ = 0 <=> a + b + c = 0

<=> Coefficient of x2 + Coefficient of y2 + Coefficient of z2 = 0

2. Planes are identical (coincident) <=> θ = 0°

<=> \(\cos \theta=1 \Leftrightarrow\left|\frac{a+b+c}{\sqrt{\left\{(a+b+c)^2+4\left(f^2+g^2+h^2-a b-b c-c a\right)\right\}}}\right|=1\)

<=> (a + b + c)² = (a + b + c)² + 4(f² +g² + h² – ab – bc – ca)

<=> f² + g² + h² – bc – ca – ab = 0 <=> (f² – bc) + (g² – ac) + (h² – ab) = 0

<=> f² = bc, g² = ac, h² = ab (∵ f² ≥ bc, g² ≥ ac, h² ≥ ab)

Note. 1. l₁x + m₁y + n₁z = 0, l₂x + m₂y + n₂z = 0 are two planes intersecting in a line with d.rs., l, m, n.

⇒ l₁l + m₁m + n₁n = 0, l₂l + m₂m + n₂n = 0 ⇒ \(\frac{l}{m_1 n_2-m_2 n_1}=\frac{m}{n_1 l_2-n_2 l_1}=\frac{n}{l_1 m_2-l_2 m_1}\)

⇒ \(\frac{l}{\left[\left(m_1 n_2-m_2 n_1\right)^2-4 m_1 n_2 n_1 m_2\right]}=\frac{m}{\sqrt{\left[\left(n_1 l_2-n_2 l_1\right)^2-4 n_1 l_2 n_2 l_1\right]}}=\frac{n}{\sqrt{\left[\left(l_1 m_2-l_2 m_1\right)^2-4 l_1 m_2 l_2 m_1\right]}}\)

⇒\(\frac{l}{\sqrt{\left(4 f^2-4 b c\right)}}=\frac{m}{\left(\sqrt{\left.4 g^2-4 a c\right)}\right.}=\frac{n}{\sqrt{\left(4 h^2-4 a b\right)}}\frac{l}{\overline{\left(f^2-b c\right)}}=\frac{m}{\left(\sqrt{\left.g^2-a c\right)}\right.}=\frac{n}{\sqrt{\left(h^2-a b\right)}}\)

⇒ d.rs. of the common line are \(\sqrt{\left(f^2-b c\right)}, \sqrt{\left(g^2-a c\right)}, \sqrt{\left(h^2-a b\right)}\)

2. If θ(≤ π/2) is the angle between the pair of intersecting planes given by ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0, then

⇒ \(\cos \theta=\left|\frac{a+b+c}{\sqrt{\left\{(a+b+c)^2+4\left(f^2+g^2+h^2-a b-b c-c a\right)\right\}}}\right|\).

d.rs. of the line of intersection are \(\sqrt{\left(f^2-b c\right)}, \sqrt{\left(g^2-a c\right)}, \sqrt{\left(h^2-a b\right)}\)

Chapter 3 The Plane Solved Problems

Example. 1. Prove that the equation 2x² – 6y² – 12z² + 18yz + 2zx + xy = 0 represents a pair of planes, and find the angle between them.

Solution.

Let the given equation be ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 …..(1)

comparing the given equation to (1), a =2, b = -6, c = -12, f = 9, g = 1, h = 1/2

∴ \(\mathrm{D}=\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{ccc}
2 & 1 / 2 & 1 \\
1 / 2 & -6 & 9 \\
1 & 9 & -12
\end{array}\right|\)

= \(2(72-81)-\frac{1}{2}(-6-9)+1\left(\frac{9}{2}+6\right)=-18+\frac{15}{2}+\frac{21}{2}=0\)

f² = 81, bc = 72 ⇒ f² > bc, ac = -24 ⇒ g² > ac, h² = 1/4, ab = -12 ⇒ h² > ab.

∴ The given equation represents a pair of planes through the origin.

Let θ be the acute angle between the planes.

∴ \(\cos \theta=\left|\frac{a+b+c}{\sqrt{\left[(a+b+c)^2+4\left(f^2+g^2+h^2-a b-b c-c a\right)\right]}}\right|\)

OR : 2x² – 6y² – 12z² + 18yz + 2zx + xy = 0

⇒ 2x² + x(y + 2z) – (6y² + 12z² – 18yz) = 0

⇒ \(x=\frac{-(y+2 z) \pm \sqrt{\left[(y+2 z)^2+8\left(6 y^2+12 z^2-18 y z\right)\right]}}{4}\)

⇒ \(4 x=-(y+2 z) \pm \sqrt{\left(49 y^2-140 y z+100 z^2\right)} \Rightarrow 4 x=-y-2 z \pm(7 y-10 z)\)

⇒ 4x – 6y + 12z = 0, 4x + 8y – 8z = 0 ⇒ 2x – 3y + 6z = 0, x + 2y – 2z = 0

∴ The given equation represents a pair of planes through the origin.

⇒ \(\cos \theta=\frac{2(1)-3(2)+6(-2)}{\sqrt{(4+9+36)} \cdot \sqrt{(1+4+4)}}=\frac{+16}{21}\)  ∴ \(\theta=\cos ^{-1}\left(\frac{16}{21}\right)\)

Example. 2. If a² + b² + c² > 2ab + 2bc + 2ca, show that the equation ax² + by² + cz² -(a + b – c)xy – (b + c – a)yz – (c + a – b)zx = 0 represents a pair of planes. Also show that the line of intersection of the planes makes equal angles with the coordinate axes.

Solution.

Given

a² + b² + c² > 2ab + 2bc + 2ca,

Let H ≡ ax² + by² + cz² -(a + b – c)xy – (b + c – a)yz – (c + a – b)zx = 0 ……(1)

∴ Determinant of H = D

=\(\left|\begin{array}{ccc}
a & \frac{-(a+b-c)}{2} & \frac{-(c+a-b)}{2} \\
\frac{-(a+b-c)}{2} & b & \frac{-(b+c-a)}{2} \\
\frac{-(c+a-b)}{2} & \frac{-(b+c-a)}{2} & c
\end{array}\right|\)

= \(-\frac{1}{8}\left|\begin{array}{ccc}
2 a & (a+b-c) & (c+a-b) \\
(a+b-c) & -2 b & (b+c-a) \\
(c+a-b) & (b+c-a) & -2 c
\end{array}\right| R_1=R_1+R_2+R_3\).

= \(-\frac{1}{8}\left|\begin{array}{ccc}
0 & 0 & 0 \\
a+b-c & -2 b & b+c-a \\
c+a-b & b+c-a & -2 c
\end{array}\right|=0\)

[f² > bc be condition]:

⇒ \(\left(\frac{b+c-a}{2}\right)^2-b c=\frac{a^2+b^2+c^2-(2 a b+2 b c+2 c a)}{4}>0 \text { (by hyp.) }\)

Similarly [g2 > ac, h2 ≥ ab conditions] \(\left(\frac{c+a-b}{2}\right)^2-a c>0,\left(\frac{a+b-c}{2}\right)^2-a b>0\)

∴ (1) represents a pair of intersecting planes.

For the common line \(\overrightarrow{\mathrm{PQ}}\) of intersection of the planes, d.rs. are

⇒ \(\sqrt{\left[\left(\frac{b+c-a}{2}\right)^2-b c\right]}, \sqrt{\left[\left(\frac{c+a-b}{2}\right)^2-c a\right]}, \sqrt{\left[\left(\frac{a+b-c}{2}\right)^2-a b\right]}\)

i.e. \(\frac{\sum a^2-2 a b}{4}, \frac{\sum a^2-2 a b}{4}, \frac{\sum a^2-a b}{4}\)

But d.cs. of \(\overrightarrow{\mathrm{OX}}, \overrightarrow{\mathrm{OY}}, \overrightarrow{\mathrm{OZ}} \text { are } 1,0,0 ; 0,1,0 ; 0,0,1\).

Let \(\theta=\left(\begin{array}{ll}
\overrightarrow{P Q} & \overrightarrow{O X}
\end{array}\right)\)

∴ \(\cos \theta=\frac{1.1+1.0+1.0}{\sqrt{3} \sqrt{1}}=\frac{1}{\sqrt{3}} \text { i.e. } \theta=\cos ^{-1} \frac{1}{\sqrt{3}}\)

Similarly we can observe that \(\overrightarrow{\mathrm{PO}} \overrightarrow{\mathrm{OY}}=\operatorname{Cos}^{-1}(1 / \sqrt{3}),(\overrightarrow{\mathrm{PQ}} \overrightarrow{\mathrm{OZ}})=(1 / \sqrt{3})\)

∴ The common line \(\overleftrightarrow{P Q}\) makes equal with the axes.

Example.3. Show that the equation x² + 4y² + 9z²– 12yz – 6zx + 4xy + 5x + 10y – 15z + 6 = 0 represents a pair of parallel planes and find the distance between them.

Solution.

Given

x² + 4y² + 9z² – 12yz – 6zx + 4xy = (x + 2y – 3z)2

∴ x² + 4y² + 9z² – 12yz – 6zx + 4xy + 5x + 10y – 15z + 6

≡ (x + 2y – 3z + k)(x + 2y – 3z – l) where

k + l = 5, 2k + 2l = 10, – 3k – 3l = -15, kl = 6 i.e., k = 3, l = 2

∴ The given equation represents the planes

x + 2y – 3z + 3 = 0, x + 2y – 3z + 2 = 0 which are parallel.

∴ Distance between the parallel planes =\(\frac{|3-2|}{\sqrt{(1+4+9)}}=\frac{1}{\sqrt{(14)}}\)

Introduction

Logical development of any branch of mathematics depends on a set. The elements of the set are underfilled terms. Associated with them are certain statements which are taken as axioms or postulates for the subject.

In earlier classes, the set language was used for the effective understanding of geometrical concepts. It may be recalled that the structure of geometry was developed by taking point, line, plane, and space as undefined concepts and that every geometrical figure is a set of points.

Solid analytical geometry (Three-dimensional coordinate geometry) is a subject redeveloped on the above lines. The study of quantitative and qualitative nature of space very much depends upon coordinates and algebraic operations and methods associated with it. Also, vectors as ordered triads have an application in the treatment of solid analytical geometry.

We take a set S and call it the 3-D space, \(R^3\) – space and its elements the points of the 3-D space. We take line L, plane π, sphere S, etc. as subsets of S.

If P ∈ L, we say that P is a point on the line L, i.e., line L passes through the point P. If P ∉ L, we say that the point is not on line L and does not pass through point P.

Points on the same line are said to be collinear and points on the same plane are said to be coplanar. If L ⊂ π, we say that L is a line in the plane π i.e., the plane π contains the line L. Further, L ⊂ π implies L is not in the plane π.

We give below, not all, but certain axioms, definitions, and theorems (without proofs) of Euclidean geometry which have wide application in solid analytical geometry.

Further, some well-known terms and concepts which are used in the development of the subject and which have to be redefined are deliberately left out as their definitions and the results involving them given out in earlier classes are true even here.

Introduction To Planes In Geometry With Examples

1. Axiom. One and only one line passes through two distinct points.

If A, B are two distinct points on L, we say that L is the one and only one line which passes through A and B. We write L as \(\overleftrightarrow{\mathrm{AB}}\).

2. Axiom. One and only one plane passes through three non-collinear points.

If π is a plane passing through three non-collinear points A, B, C we say that π is the one and only one plane determined by A, B, C, We write the plane π as

⇒ \(\overleftrightarrow{\mathrm{ABC}}\).

3. Axiom. If two distinct points lie in a plane then the line through the two points lies in the plane.

If A, B are two distinct points in a plane π, we say that \(\overleftrightarrow{A B}\) lies in the plane π.

4. Axiom. If two planes intersect, they intersect in one and only one line π₁,π₂ are two intersecting planes intersecting in the line L.

5. Axiom. Every line consists of at least two points. Every plane consists of at least three non-collinear points. In space, there always exist at least four points. which are non-coplanar.

6. Axiom. Corresponding to any two points, there exists always a unique real number called the distance between the points.

If A, B are any two points, there exists a unique real number d (AB) or AB called the distance between the points A and B with the following properties (1) AB ≥ 0, (2) AB = 0 <=> A = B (3) AB = BA (4) AC + CB ≥ AB where C is any point.

7. Definition. If A, P, B are three collinear points such that AP+PB=Ab then we say that P lies between A and B on the line. We write A-P-B. The set of points P is called the line segment between A and B with endpoints A,B. we denote the line segment as AB.

It is to be understood that the meaning of AB is to be understood depending on the context.

If A-P-B and Ap=PB, then P is called the middle point of the line segment AB.

If A, B are different points, then AB ∪ {P/A-B-P} is called the ray from A through B and we write it as \(\overrightarrow{\mathrm{AB}}\)

8. Definition. Two lines \(L_1, L_2\) in a plane π are said to intersect if \(L_1 \cap L_2 \neq \phi\) Now \(L_1, L_2\) are called intersecting lines.

9. Definition. Two lines \(L_1, L_2\) in a plane π are said to be parallel if either \(L_1, L_2\) are coincident \(\left(L_1=L_2\right) \text { or } L_1, L_2\) are not intersecting \(\left(L_1 \cap L_2 \neq \phi\right)\)

We write \(\mathrm{L}_1 \| \mathrm{L}_2\).

If A,B ∈ \(L_1\) and C,D ∈ \(L_2\) then write

⇒ \(\overleftrightarrow{A B} \| \overleftrightarrow{C D}\) or \(\mathrm{AB} \| \mathrm{CD}\).

If \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{CD}}\)

are non-collinear, \(\overleftrightarrow{A B} \| \overleftrightarrow{C D}\)

and B, D lie on the same side of \(\overleftrightarrow{\mathrm{AC}}\)

we say that \(\overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{CD}}\).

Also if \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{CD}}\)are collinear we say that

⇒ \(\overleftrightarrow{\mathrm{AB}} \| \overleftrightarrow{\mathrm{CD}}\).

If \(L_1, L_2\) are not parallel we write \(\mathrm{L}_1 / / \mathrm{L}_2\).

Definition Of A Plane In Euclidean Geometry

10. Axiom. To a given line, through a given point one and only one parallel line exists.

11. Theorem. Two distinct lines cannot intersect at more than one point.

12. Theorem. If L is a line not in the plane π and intersecting π, then the line L intersects the plane π in one and only one point.

If the point is M in π, then M is called the foot of L in π.

13. Theorem. A plane containing a line and a point not on the line is unique.

If L ⊂ π and P ∉ L, then π is unique.

14. Theorem. Two intersecting lines determine a unique plane.

If \(\mathrm{L}_1 \text { and } \mathrm{L}_2\) are two intersecting lines \(\mathrm{L}_1 \text { and } \mathrm{L}_2\) determine a plane.

Axioms Of Planes In Euclidean Geometry Explained

15. Axiom. For every angle, there corresponds a unique number between 0 and π. This real number is called the measure of the angle.

The measure of an angle when measured by comparison with selected standard radial, the measure will be in radians. When the selected standard-a right angle is, the measure will be in degrees.

If θ is the angle between \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\), we write θ = \(\angle(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}})=(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}})(\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{AB}})=\angle \mathrm{BAC}=\angle \mathrm{CAB}\) such that 0 ≤ θ ≤ π.

Also \((\overrightarrow{A B}, \overrightarrow{A B})=0\) and \((\overrightarrow{A B}, \overrightarrow{B A})=\pi\).

If \(\angle \mathrm{BAC}=\frac{\pi}{2}\),we say that\(\overrightarrow{\mathrm{BA}}\) is perpendicular to \(\overleftrightarrow{\mathrm{AC}}\)and we write

⇒ \(\overleftrightarrow{\mathrm{BA}} \perp \overleftrightarrow{\mathrm{AC}}\) or

⇒ \(\overleftrightarrow{\mathrm{AB}} \perp \overleftrightarrow{\mathrm{AC}}\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}}\) or AB ⊥ Ac.

If \(\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}}\) are parallel to two lines

⇒ \(\mathrm{L}_1, \mathrm{~L}_2\) respectively, the angle between L₁ and L₂ is \((\overrightarrow{O P}, \overrightarrow{O Q})\)

or \(\pi-(\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}})\).

Angle between parallel lines is 0 or π.

Here π is the measure of an angle and not a symbol used to denote a plane and the meaning of π is to be understood depending on context.

16. Definition. A line L cuts (or intersects) a plane π in a point P. If all the lines in π through P are perpendicular to L, then the line L is said to be perpendicular to the plane π. We write L ⊥ π or π ⊥ L. If M ∈ L, then we also write \(\overleftrightarrow{\mathrm{PM}} \perp \pi\) or PM ⊥ π.

17. Theorem. L₁ and L₂ are two lines intersecting at P. then L is perpendicular to the lines L₁ and L₂ at P is perpendicular to the plane determined by L₁ and L₂.

18. Theorem. P is a point and L is a line. Through P and perpendicular to L one and only plane (π) exists.

If L meets π in M, the PM ⊥ L.

19. Theorem. The set of points so that each point is equidistant from two points A and B determine a unique plane (π) perpendicular to AB and intersecting AB at its middle point(C).

If P, Q , R… are points such that AP=PB, AQ=QB, AR=RB, …., then P, Q, R… are in the plane π where π ⊥ Ab at C

20. Theorem. If L₁,L₂ are two distinct lines perpendicular to the plane π, then L₁ and L₂ are parallel and coplanar.

21. Theorem. P is a point in the plane π. One and only one perpendicular line to π exists through P.

22. Theorem. L₁,L₂ are two parallel lines. If a plane is perpendicular to L₁ then it is also perpendicular to L₂.

23. Theorem. one and only one perpendicular line can be drawn to a plane from a point, not in the plane.

24. Theorem. L is a line and P is a point on L. If π is the plane perpendicular to L at P, then all the lines perpendicular to L through P lie in the plane π.

25. Definition. L is a line and π is a plane. If no point of L is in π(L∩π=ϕ) or if L is in π, then L is said to be parallel to π and we write L ∥ π.

26. Definition.  π₁, π₂ are two planes which are either coincident (π₁=π₂) or parallel (π₁ ∩ π₂ = ϕ). Then π₁,π₂ are said to be parallel and we write π₁ ∥ π₂.

27. Theorem. π₁,π₂ are two distinct parallel planes. If a plane π cuts π₁ in a line L₁ and π₂ in a line L₂, then L₁∥ L₂.

28. Theorem. If L is a line perpendicular to the plane π₁, then L is perpendicular to the planes parallel to π₁.

29. If π₁, π₂ are two planes perpendicular to the lines L, then π₁ ∥ π₂.

30. Theorem. If L₁ is a line parallel to any line in a plane π then L₁ ∥ π.

31. Theorem. L₁ is a line parallel to the plane π₁. π₂ is a plane containing L₁ and intersecting π₁ in L₂. Then L₁ ∥ L₂.

32. Theorem. L is a line in a plane π. R is a point on L. Also Q is a point in π, but not on L. Further, P is a point not in π. Then

1) QR ⊥ L and PR ⊥ L ⇒ PQ ⊥ π
2) PR ⊥ L and PQ ⊥ π ⇒ QR ⊥ L
3) PQ ⊥ π and QR ⊥ L ⇒ PR ⊥ L
This is called the theorem of three perpendiculars.

33. Theorem. Two parallel planes π₁, π₂ always make equal intercepts or lines perpendicular to π₁,π₂.

Definition. L is perpendicular to π₁, π₂. The intercept on L by π₁,π₂ is called the distance between π₁ and π₂.

34. Definition. If L₁, L₂, L₃ are three lines such that L₁ ∩ L₂ ∩ L₃ = P, then the lines L₁, L₂ and L₃ are said to be concurrent.

35. Theorem. If L₁ ∥ L₂, L₂ ∥ L3, then L₁ ∥ L₃.

We say that L₁, L₂, L₃ are parallel and we write L₁ ∥ L₂ ∥ L₃.

Examples Of Planes In 2d And 3d Euclidean Geometry

36. Theorem. π₁, π₂, π₃ are three distinct planes so that no two of which are parallel. The three lines of their intersection are either parallel or concurrent.

37. Projections. Here projection will mean orthogonal projection only.

1. Definition. P is a point and L is a line. P ∉ L. If M is the foot of the perpendicular from P on L, then M is called the projection of P in L.

2. Definition. P is a point and π is a plane not containing P. If the perpendicular from P to π meets π in the point M, then M is called the projection of P in π.

3. Definition. P, Q are two points and L is a line, P ∈ L and Q ∉ L. If M is the projection of Q and L, the PM is called the projection of PQ in L.

4. Definition. P, Q are two points and L is a line. P, Q ∉ L. If M,N are the respective projections of P, Q in L, then the line segment MN is called the projection of the line segment PQ in L.

5. Theorem. π is a plane and L is a line not in π. If perpendicular are drawn from every point on L to π, then all the projections in π are either collinear (when L is not perpendicular to π) or coincident in a point (when L ⊥ π).

38. Some Useful results

1. L is a line and P is a point on it. The perpendicular to L at P are coplanar.
2. The planes perpendicular to a line are all parallel.
3. L is a line perpendicular to a plane π. All the planes through L are perpendicular to π.
4. L₁, L₂ are a pair of intersecting lines. L₃, L₄ are a second pair of intersecting lines so that L₃ ∥ L₁ and L₄ ∥ L₂.

Then

1. the angle between the first pair=the angle between the second pair.
2. the plane determined by the first pair is parallel to the plane determined by the second pair.

39. Definition. L is a line and R is the set of real numbers. \(f: L \rightarrow R\) is a one-one mapping. If A,B ∈ L such that \(|A, B|=|f(A)-f(B)|\), then f is called a coordinate system for L.

The real number x(=f(p)) is called the coordinate of P w.r.t the coordinate system f on the line L and we write it as p(x).

In this context, L is called the coordinate line. The set p(x) is called the geometric figure on L.

Axiom. Every line has a coordinate system.

A coordinate system on the line L depends on arbitrarily chosen points O and I on L such that

1. O, called the origin, corresponds to the number of the O, and
2. I, called the unit point, to correspond to the number 1.

Clearly, we can have infinitely many coordinate systems defined on L.

Then for every real number x and for each point P on L one-to-one correspondence exists with the following properties and notation,p(x) denotes the point p with coordinate x’

p(a),Q(b) ∈ L ⇒

1. p(a) = 0 <=> p=0,
2. p(a) lies to the right of O is a>0, and Q(b) lies to the left of O if b <0,
3. Between P and Q, distance = PQ=|b-a| and directed distance = b-a,
4. p(a) lies to the right of Q(b) if a > b,
5. Q(b) lies to the left of p(a) if b < a,
6. p(A) coincides with Q(b) i.e., P=Q if a=b.

40. Definition. A(a)., P(x), B(b) are points on the coordinate line L. Then

1. A-P-B ⇒ is said to divide the line segment AB internally in the ratio (x-a):(b-x) and x-a,b-x are positive.
2. P-A-B or A-B-P ⇒ is said to divide the line segment AB externally in the ratio (x-a,(b-x) are of opposite signs. we write (P;A,B)=(x-a):(b-x). It is +ve or -ve according as P divides AB internally or externally.

IF (P;A,B) = m:n, then (x-a):(b-x) = m:n ⇒ n(x-a)=m(b-x)

41. Theorem. In a given ratio a line segment is divided at one an only the point.

42. Skew lines.

Definition. Any two non-parallel and non-intersecting lines are called skew lines.

Since any two lines in a plane must be either parallel or intersecting, skew lines are non-coplanar. Conversely, any two non-coplanar lines are skew lines.

43. Theorem. L₁,L₂ are two skew lines. Then there exist one and only one plane π through one of the lines and parallel to the second.

44. Theorem. If \(\overleftrightarrow{\mathrm{MN}}\) is the projection of a line L in a plane π, then \(\overleftrightarrow{\mathrm{MN}}\), L are coplanar.

If L ∥ π, then L ∥ \(\overleftrightarrow{\mathrm{MN}}\).

45. Theorem. If L₁,L₂ are two skew lines, then there exists one and only one line that intersects L₁,L₂ and is perpendicular to L₁,L₂.

Properties and axioms of planes in mathematics

46. Definition. L is a line and π₁ is a plane not containing L. If L is not parallel to π₁, then the angle between L and π1 (written as (L, π₁)) is the angle between L and \(\overleftrightarrow{\mathrm{MN}}\), where \(\overleftrightarrow{\mathrm{MN}}\) is the projection of L in π₁.

We write \(\left(\mathrm{L},\pi_1\right)=(\mathrm{L}, \overleftrightarrow{\mathrm{MN}})\).

We can have (L,π₁)= \(\frac{\pi}{2} \pm \theta\)

where θ is the acute angle between L and a normal to π₁.

The Sphere Angles Of Intersection Of Spheres, Orthogonal Spheres

Definition Of A Sphere In Geometry With Examples And Proofs

Definition. P is a common point to two spheres ξ₁, ξ₂. Any angle θ between the tangent planes at P to two spheres is called an angle of intersection of the spheres ξ₁, ξ₂ at P. The other angle between the spheres π – θ.

If θ = π/2 the spheres are said to interest orthogonally at P and the spheres are called orthogonally spheres.

Theorem.1  ξ₁, ξ₂ are two intersecting spheres (not touching). r₁,r₂ are their respective radii and d is the distance between their centres. If P is a common point to ξ₁, ξ₂ then an angle θ of the intersection of the spheres ξ₁, ξ₂ at P is given by

⇒ \(\cos \theta=\pm\left(\frac{r_1^2+r_2^2-a^2}{2 r_1 r_2}\right)\)

Proof. Let A, B be the centres of the spheres ξ₁, ξ₂. The tangent planes at P to ξ₁, ξ₂ are perpendicular to AP, Bp respectively. Since the angle between the planes is the angle between their normals, ∠APB = θ or π – θ.

Ap =r₁, Bp = r₂ and AB = d.

From △APB,

AB² = AP² + PB² – 2 AB.PB cos∠APB

i.e.,d₂ = r₁² + r₂² ± 2r₁r₂cosθ

i.e., \(\cos \theta=\pm \frac{r_1^2+r_2{ }^2-d^2}{2 r_1 r_2}\)

Note 1.  Since the value of cosθ is independent of P, then angle between two spheres ξ₁, ξ₂ can be found to be the same at any point of their intersection.
2. Spheres ξ₁, ξ₂ cut orthogonally <=> θ = 90° <=> r₁² + r₂² = d²       

In this case, the tangent plane to ξ₁ at P passes through the centre of ξ₂ and the tangent plane to ξ₂ at P passes through the centre of ξ₁.

Theorem.2  S ≡ x² + y² + z² + 2ux + 2vy + 2wz + d = 0,s’ ≡ x² + y² + z² + 2u’x + 2v’y + 2w’z + d’ = are two orthogonal spheres <=> 2uu’ + 2ww’ = d + d’.

Proof. Let A, B be the centres and r1, r2 be the radii of the spheres s = 0, s’ = 0.

Spheres s=0, s’=0 cut orthogonally

⇔AB² = r₁² + r₂²

⇔ (u’-u)² + (v’-v)² + (w’-w)2 = u² + v² + w² – d + u’² + v’² + w’² – d’

⇔ -2uu’ – 2vv’ – 2ww’ = d + d’ <=> 2uu’ + 2vv’ + 2ww’ = d + d’

Theorems Related To Spheres With Solved Problems Step-By-Step

Theorem.3  If r₁, r₂ are the radii of two orthogonal spheres, then the radius of the circle of their intersection is \(\frac{r_1 r_2}{\sqrt{r_1^2+r_2^2}}\)

Proof. A, B are the centres of the two orthogonal spheres. M is the centre and a is the radius of the circle common to the spheres.

∴ A, M, B are colinear and MP ⊥ AB.

P is a common point of intersection of the spheres,

AP = r₁, BP = r₂, ∠APB = 90° ⇒ AB² = r₁² + r₂²

⇒ (AM + MB)² = r₁² + r₂²

⇒ AM² + MB² + 2 AM . MB = r₁² + r₂²

⇒ \(r_1^2-a^2+r_2^2-a^2+2 \sqrt{\left(r_1^2-a^2\right)\left(r_2^2-a^2\right)}=r_1^2+r_2^2\)

⇒ 4(r₁² – a²)(r₂² – a²) = 4a⁴

⇒\(r_1^2 r_2^2-a^2\left(r_1^2+r_2^2\right)=0 \Rightarrow a=\frac{r_1 r_2}{\sqrt{r_1^2+r_2^2}}\)

Note. If S=0, S’=0 are the equations to the orthogonal spheres with radii r₁, r₂ then 2uu’+2vv’+ww’=d+d’ and

r₁² = u² + v² + w² – d, r₂² = u’² + v’² + w’² – d’

∴ r₁² + r₂² = u² + v² + w² + u’² + v’² + w’² – (2uu’ + 2vv’ + 2ww’)

= (u-u’)² + (v-v’)² + (w-w’)²

∴ \(a=\frac{\sqrt{u^2+v^2+w^2-d} \cdot \sqrt{u^{\prime 2}+v^2+w^2-d^{\prime}}}{\left(u-u^{\prime}\right)^2+\left(v-v^{\prime}\right)^2+\left(w-w^{\prime}\right)^2}\)

The Sphere Power Of A Point

Definition. B is a point on a line L intersecting a sphere ξ with centre C and radius = a in P, Q. Then the power of the point B w.r.t. the sphere ξ is

(1) BP. BQ if B is an external point to ξ
(2) -BP.BQ if B is an internal point to ξ
(3) 0 if B is on ξ.

If B = \(\bar{b}\) (x₁, y₁, z₁) and the equation to the sphere ξ is S, then the power of the point B w.r.t ξ is S₁₁.

i.e., x₁² + y₁² + z₁² + 2ux₁ + 2vy₁ + 2wz₁ + d

= \(\overline{\mathrm{CB}}^2-a^2=\mathrm{CB}^2-(\text { radius of } \xi)^2\)

If B is an external point to ξ and \(\overleftrightarrow{\mathrm{BT}}\) is a tangent line to ξ at T, then BT² = S₁₁ = Power of the point B w.r.t ξ.

Note that: Power of the point B is independent of the d.cs.l,m,n.

Example. If the powers of a point w.r.t two given spheres are in a constant ratio, show that the locus of the point is a sphere.

The Sphere Radical Plane

Definition. The locus of points each of whose powers w.r.t two non-concentric spheres are equal is a plane called the radical plane (R.P.) of the two spheres.
ξ, ξ’ are two non-concentric spheres and π is their radical plane.
P is a point on the radical plane π. <=> power P w.r.t. ξ₁= Power of P w.r.t ξ₂.

Theorem.4 Equation to the radical plane of spheres S=0, S’=0 is S-S’=0.

Proof. S ≡ x² + y² + z² + 2ux + 2vy + 2wz + d =0

S’ ≡ x² + y² + z² + 2u’x + 2v’y + 2w’z + d’ = 0

∴ (-u, -v, -w) ≠ (-u’, -v’, -w’)

B(x1, y1, z1) is a point whose powers w.r.t the spheres are equal.

⇔ S₁₁ = S’₁₁

⇔ x₁² + y₁² + z₁² + 2ux₁ + 2vy₁ + 2wz₁ + d = x₁² + y₁² + z₁² + 2u’x₁ + 2v’y₁ + 2w’z₁ + d’

⇔ 2(u-u’)x₁ + 2(v-v’)y₁ + 2(w-w’)z₁ + (d-d’) = 0

∴ Locus of B is 2(u-u’)x+2(v-v’)y+2(w-w’)z+(d-d’)=0 Which is a plane.

But the locus of B is the radical plane of the spheres S=0, S’=0.

∴ The radical plane of the spheres S=0, S’=0 is

2(u-u’)x + 2(v-v’)y + 2(w-w’)z + d – d’ = 0 i.e., S-S’=0

Note. D.rs. of the line of centres of the spheres are u-u’, v-v’, w-w’.

∴ Radical plane is perpendicular to the line of centres of the spheres.

Note 1. The line of centres and it is perpendicular to the radical plane.

∴ Radical plane is perpendicular to the line of centres of the spheres.

2. If two spheres interest the plane of their circle of intersection is their radical place.

3. If two spheres touch, their radical plane is the tangent plane at the point of contact to either of the spheres.

The Sphere  Radical Line

Definition. If ξ, ξ’, ξ” are three spheres with non-collinear centres then the three radical planes of the spheres taken in pairs pass through a unique line, called the radical line.

Orthogonal Spheres Examples And Their Geometric Properties

Theorem.5  S=0, S’=0, S”=0 are three spheres whose centres are non-collinear, then the three radical planes of the spheres taken in pairs pass through a unique line.

Proof. Let A, B, C be the centres of the spheres S=0, S’=0, S”=0.

The radical plane (π) of S=0, S’=0 is S-S’=0 and is perpendicular to AB.

The radical plane (π’) of S’=0, S”=0 is S’-S”=0 and is perpendicular to BC.

The radical plane (π”) of S”=0, S=0 is S”-S=0 and is perpendicular to AB.

Since lines \(\overleftrightarrow{\mathrm{AB}}, \overleftrightarrow{\mathrm{BC}}\) intersect, the planes π,π’ have a line, say L in common.

All the points on L=0 lie on the plane (S-S’)+1(S’-S”)=0

i.e. S-S” = 0 i.e., S”-S=0 i.e. L lies in the plane S”-S=0

∴ π,π’,π” pass through the line L.

Note. L is the radical line of the spheres S=0, S’=0, S”=0 and its equation is S-S’=0, S’-S”=0 i.e. S=S’=S”.

The Sphere Radical Centre

Definition. The four radical lines of four spheres with non-coplanar centres, taken three by three intersect at a unique point, called the radical centre of the spheres.

Theorem.6  S=0, S’=0, S”=0, S'”=0 are four spheres whose centres are non-coplanar, then the four radical lines of four spheres taken three by three intersect at a unique point.

Proof. The radical plane of S=0, S’=0 is S-S’=0

i.e. 2(u-u’)x + 2(v-v’)y + 2(w-w’)z + (d-d’) = 0           …..(1)

The radical planes of S=0, S”=0 is S-S”=0

i.e. 2(u-u”)x + 2(v-v”)y + 2(w-w”)z + (d-d”) = 0        …..(2)

The radical planes of S=0, S”‘=0 is S-S'”=0

i.e. 2(u-u'”)x + 2(v-v'”)y + 2(w-w'”)z + (d-d'”) = 0     …..(3)

Since the centres (-u, -v, -w), (-u’, -v’, -w’),(-u”, -v”, -w”), (-u'”, -v'”, -w'”) are non-coplanar.

⇒ \(\left|\begin{array}{cccc}
-u & -v & -w & 1 \\
-u^{\prime} & -v^{\prime} & -w^{\prime} & 1 \\
-u^{\prime \prime} & -v^{\prime \prime} & -w^{\prime \prime} & 1 \\
-u^{\prime \prime \prime} & -v^{\prime \prime \prime} & -w^{\prime \prime \prime} & 1
\end{array}\right| \neq 0\)

⇒ \(\left|\begin{array}{cccc}
-u & -v & -w & 1 \\
u-u^{\prime} & v-v^{\prime} & w-w^{\prime} & 0 \\
u-u^{\prime \prime} & v-v^{\prime \prime} & w-w^{\prime \prime} & 0 \\
u-u^{\prime \prime \prime} & v-v^{\prime \prime \prime} & w-w^{\prime \prime \prime} & 0
\end{array}\right| \neq 0\)

⇒ \(\left|\begin{array}{ccc}
u-u^{\prime} & v-v^{\prime} & w-w^{\prime} \\
u-u^{\prime \prime} & v-v^{\prime \prime} & w-w^{\prime \prime} \\
u-u^{\prime \prime \prime} & v-v^{\prime \prime \prime} & w-w^{\prime \prime \prime}
\end{array}\right| \neq 0\)

⇒ Radical planes (1), (2), (3) pass through a unique point P.

∴ P lies on the radical planes of S=0, S’=0, S”=0, S'”=0,

⇒ P lies on the radical line of S=0, S’=0, S”=0

Similarly P lies on the radical line of S=0,S’=0,S”‘=0,

P lies on the radical line of S=0, S”=0, S”‘=0.

P lies on the radical line of S’=0,s”=0,S”‘=0.

∴ The four radical lines of four spheres taken three by three intersect at the unique point P.

Note. P is the radical centre of the spheres S=0, S’=0, S”=0, S'”=0.

Step-By-Step Solutions For Sphere Geometry Problems

Theorem.7  The centre of the sphere ξ which intersects two spheres ξ₁, ξ₂ orthogonally lies on the radical plane of the sphere ξ₁, ξ₂.

Proof. Let A, B be the centres and r₁, r₂ be the radii of the spheres ξ₁, ξ_2.

Let C be the centre of the sphere ξ. Let T₁, T₂ be the respective common points to ξ₁, ξ₂  and ξ₂, ξ.

ξ intersects ξ₁ orthogonally

⇒ CT₁ ⊥ AT₁ ⇒ AC²-AT₁² = CT₁²

⇒ AC² – (radius of ξ₁)² = CT₁² Art.6.32

⇒ Power of the Point C w.r.t ξ₁ = CT₁² =(radius of ξ)².

Similarly power of point C w.r.t ξ₂ = CT₂² =(radius of ξ)².

∴ Power of the point C w.r.t ξ₁ = Power of the point C w.r.t ξ₂.

∴ C lies on the radical plane of ξ₁, ξ₂.

OR
Let S’=0, S”=0 be the equations of the spheres ξ₁, ξ₂ and S=0 be the equation of ξ.

Radical plane of S’=0, S”=0 is S’-S”=0

i.e. 2(u’-u”)x + 2(v’-v”)y + 2(w’-w”)z + d’-d” = 0

ξ intersects ξ₁ orthogonally ⇒ 2uu’ + 2vv’ + 2ww’ = d + d’        …..(1)

ξ intersects ξ₁ orthogonally ⇒ 2uu” + 2vv” + 2ww” = d + d”     …..(2)

(1)-(2): 2(u’-u”)x + 2(v’-v”)y + 2(w’-w”)z + d’ – d” = 0

⇒ 2(u’-u”)(-u) + 2(v’-v”)(-v) + 2(w’-w”)(-w) + d’ – d” = 0

∴ The centre (-u, -v, -w) of S=0 clearly lies on the radical plane of s’=0, s”=0.

i.e. the centre of ξ lies on the radical plane ξ₁, ξ₂.

The Sphere Solved Problems

Example.1. Find the equation of the sphere through the circle x² + y² + z² – 2x + 3y -4z +6 = 0 , 3x-4y+5z-15=0 and cutting the sphere x² + y² + z² + 2x + 4y -6z +11 = 0 orthogonally.

Solution:

Given

x² + y² + z² – 2x + 3y -4z +6 = 0 , 3x-4y+5z-15=0

x² + y² + z² + 2x + 4y -6z +11 = 0

Let a sphere through the given circle and cut the sphere

x² + y² + z²  + 2x + 4y -6z +11 = 0 …..(1) orthogonally be

x² + y² + z²  – 2x + 3y -4z + 6 + λ(3x-4y+5z-15)=0

i.e., x² + y² + z² + (3λ-2)x + (-4λ+3)y + (5λ-4)z + (-15λ+6) = 0

∴ \(\frac{2(3 \lambda-2)}{2}+\frac{4(-4 \lambda+3)}{2}-\frac{6(5 \lambda-4)}{2}=11-15 \lambda+6 \quad \text { i.e. } \lambda=-1 / 5\)

∴ Equations to the required sphere is 5(x² + y² + z² ) – 13x + 19y – 25z + 45 = 0.

Radical Planes Of Spheres Explained With Solved Examples

Example. 2. Find the equation of the sphere which touches the plane 3x + 2y – z + = 0 at (1, -2, 1) and cuts orthogonally the sphere x² + y² + z² – 4x + 6y + 4 = 0.

Solution:

Given

3x + 2y – z + = 0 and (1, -2, 1)

x² + y² + z² – 4x + 6y + 4 = 0

For the sphere x² + y² + z² – 4x + 6y + 4 = 0 …..(1)

centre = (2, -3, 0), radius = \(\sqrt{4+9-4}=3\)

Since the plane, 3x + 2y – z + 2 = 0 at (1, -2, 1) is the tangent plane to the required sphere, equations to the normal at (1, -2, 1) is the tangent plane to the required sphere, equations to the normal at (1, -2, 1) are

⇒ \(\frac{x-1}{3}=\frac{y+2}{2}=\frac{z-1}{-1} \text { (= } t \text { say) }\)

∴ Centre of the required sphere can be taken as (3t+1, 2t-2, -t+1)

and radius = \(\sqrt{(3 t+1-1)^2+(2 t-2+2)^2+(-t+1-1)^2}=\sqrt{14}|t|\)

Since the required sphere cuts orthogonally (1), (3t+1-2)² + (2t-23)² + (-t+1-0)² = 14t² + 9 i.e., t = -3/2.

∴ Centre of the required sphere \(\left(-\frac{7}{2},-5, \frac{5}{2}\right)\)

∴ Equation to the required sphere is \(\left(x+\frac{7}{2}\right)^2+(y+5)^2+\left(z-\frac{5}{2}\right)^2=\left(\frac{3}{2} \times \sqrt{14}\right)^2\)

i.e. x² + y² + z² + 7x + 10y – 5z + 12 = 0.

Example. 3. Find the radical centre of the sphere

x² + y² + z² + 4y = 0                            …..(1)
x² + y² + z² + 2x + 2y + 2z + 2 = 0    ….(2)
x² + y² + z² + 3x – 2y + 8z + 6 = 0     ….(3)
x² + y² + z² – x + 4y – 6z – 2 = 0         ….(4)

Solution: R.P. of (1) and (2) is 2x – 2y + 2z + 2 = 0

i.e., x-y+z+1=0                                 ……(5)

R.P. of (1) and (3) is 3x-6y+8z+6=0 ……(6)

R.P. of (1) and (4) is x+6z+2 = 0      ……(7)

R.P. of (3) and (4) is 4x-6y+14z+8=0 i.e.,

2x-3y+7z+4=0         ……(8)

∴ Radical line of the spheres (1), (2), (3) is

x – y + z +1 = 0                                                             …..(5)

3x – 6y + 8z + 6 = 0                                                      …..(6)

and radical line of the spheres (1),(3),(4) is x+6z+2=0 …..(7)

2x – 3y + 7z + 4 = 0                                                     …..(8)

The point of the intersection of these radical lines is the radical centre of the spheres.

3 x (5)-(8) : x – 4z -1 = 0                                                ……(9)

(7)-(9) : 10z + 3 = 0 ⇒ \(z-\frac{3}{10}\)

∴ \(x=-\frac{1}{5}, y=\frac{1}{2}\)

∴ Radical centre of the spheres = \(\left(-\frac{1}{5}, \frac{1}{2},-\frac{3}{10}\right)\).

Solved Exercises On Spheres And Radical Planes In Geometry

Example.4. Show that all spheres through the origin and each set of points where the planes parallel to the plane \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\) cut the axes, from a system of spheres which are cut orthogonally by the sphere x² + y² + z² + 2fx + 2gy + 2hz = 0 …(1) if af + bg + ch = 0.

Solution:

Given

x² + y² + z² + 2fx + 2gy + 2hz = 0 …(1) if af + bg + ch = 0

Let a plane parallel to

⇒ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\) be \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\lambda(\lambda \neq 0)\)

Let it meet the axes A, B, C.

∴ A=(λa,0,0), B=(0,λb,0), C=(0,0,λc),

∴ The equation to the sphere through O, A, B, C is x² + y² + z² – λax – λby – λcz = 0

This intersects (1) orthogonally if

⇒ \(2 \cdot f \cdot\left(\frac{-\lambda a}{2}\right)+2 g\left(\frac{-\lambda b y}{2}\right)+2 h\left(\frac{-\lambda c z}{2}\right)=0\)

i.e., af + bg + ch = 0 (∵ λ≠0).

The Sphere  Coaxal System Of Spheres

Definition. A system of spheres such that any two spheres of the system have the same radical plane is called a coaxal system of spheres.

S=0 and S’=0 are two spheres of a coaxal system of sphere ξ.

⇒ S-S’=0 is the radical plane of the coaxal system of sphere ξ.

Theorem.8 If S=0 is a sphere and U=0 is a plane, then the equation s+λU=0(λ being real) represents a coaxal system of spheres with radical plane U=0.

Proof. Given that S=0 is a sphere and U=0 is a plane

Consider the equation S+λU=0, A being real ….(1)

Let S + λ₁U=0                          ……(2) and

S + λ₂U=0                               ……(3)

(λ₁ ≠ λ₂) be two spheres of the system (1).

Radical plane of(2) and (3) is (λ₁ ≠ λ₂), U=0

∴ U=0 (∵ λ₁ ≠ λ₂), independent of λ.

∴ Every two spheres of the system (1) have the same radical plane U=0.

∴ S+λU=0 is the equation to coaxal system of spheres with radical plane U=0.

Note. S=0, and S’=0 are two non-concentric spheres. Then S-S’=0 is the radical plane of S=0, S’=0.

∴ S+λ(S-S’)=0, being real, is a coaxal system of spheres

i.e. (1+λ)S+(-λ)S’=0 is a coaxal system of spheres

i.e. λ₁S+λ₂S’=0, (λ₁,λ₂)≠(0,0) and λ₁+λ₂≠0 represents a coaxal system of spheres with radical plane S-S’=0.

Theorem.9 The centres of the spheres of a coaxal system of spheres are collinear.

Proof. Let ξ be a coaxal system of spheres with radical plane π.

Let ξ₁, ξ₂ be two spheres of the system ξ with centres A, B.

∴ AB ⊥ π. …..(1)

Let ξ₃ be a sphere of the system ξ distinct from ξ₁, ξ₂ with centre C.

∴ ξ₁, and ξ₃ have the same radical plane and AC ⊥ π. …..(2)

∴ From (1) and (2), A, B, C are collinear.

∴ All the centres of the spheres of the system lie on \(\overleftrightarrow{\mathrm{AB}}\)

i.e. all the centres are collinear.

⇒ \(\overleftrightarrow{\mathrm{AB}}\) is called the line of centres of the coaxal system ξ.

Note. The radial plane of a coaxal system of spheres is perpendicular to the line of centres of the system.

The Sphere A Simplified Form Of The Equation To A Coaxal System Of Spheres.

Theorem.10 A Coaxal system of spheres can be reduced to the form x² + y² + z²+ 2λx + d = 0 where d is a constant and λ is a parameter.

Proof. The line of centres of a coaxal system of spheres is perpendicular to the radical plane of the system.

Let the point of intersection of the line of centres with the radical plane be origin O, the line of centres be X-axis and the radical plane be YZ plane.

With this frame of reference, let a sphere of the coaxal system be

x² + y² + z²  + 2ux + d = 0 ……(1) (∵ centre lies on the x-axis)

where u, and d are parameters. O is a point on the radical plane of the system.

∴ Power of O w.r.t. (1) = 0 + 0 + 02u(0) + d = d.

Since the power of the point O w.r.t. any sphere of the system must be the same, d is a constant.

∴ Equation to the coaxal system of spheres can be taken as x² + y² + z² + 2λx + d = 0 where λ(=u) is a parameter and d is a constant.

Note. The equation x² + y² + z² + 2λx + d = 0 (λ is a parameter and d is a constant) represents a coaxal system with the line of centres as X-axis and the radical plane as YZ plane.

By giving values to λ and taking f as constant, we get spheres of the coaxal system.

Consider a coaxal system of spheres x² + y² + z² + 2λx + d = 0 …….(1)

where d is a constant and λ is a parameter.

The radical plane of the system is the YZ plane i.e. x = 0 …..(2)

∴ (1) and (2) intersect in points given by x = 0,  y² + z² + d = 0

i.e.  y² + z² = -d.

(1) d < 0.

All the points of intersection lie on the circle x = 0,\(y^2+z^2=(\sqrt{-d})^2\) and every sphere of the system passes through the circle.

∴ The coaxal system is an intersecting type of coaxal system of spheres.

(2) d = 0.

∴ x=0, y² + z²= 0 i.e. x = 0, y = 0, z = 0

i.e. (0, 0, 0) is a point common to (1) and (2).

i.e. every pair of spheres of the system touch at (0, 0, 0) and the radical plane of the system is the system is the tangent plane at (0, 0, 0) to every sphere of the system.

∴ The coaxal system is a touching type of coaxal system such that every pair touches at (0,0,0).

(3) d > 0.

∴ There are no points common to (1) and (2)

i.e. there are no points in common to any two spheres of the system.

∴ The coaxal system is a non-intersecting type of coaxal system of spheres.

The Sphere Limiting Points

Definition. Point spheres of a coaxal system of spheres are called limiting points of the system.

Let x² + y² + z² + 2λx + d = 0 where d is a constant and λ is a parameter, represent a coaxal system of spheres.

For any sphere of the system, radius = \(\sqrt{\left(\lambda^2-d\right)}\) and centre = (-λ, 0, 0)

For limiting points of the system, radius = 0

i.e. \(\sqrt{\left(\lambda^2-d\right)}=0\) i.e. λ = ±√d

(1) If d=0 then λ=0 and hence the system has only one limiting point and it is, (0, 0, 0). In this case the system is a touching type of coaxal system of spheres at (0, 0, 0).

(2) If d>0; then λ has two values ±√d and hence the system has two limiting points only. The limiting points are (√d, 0, 0), (-√d, 0, 0). In this case, no two spheres of the system intersect.

(3) If d<0, the system has no limiting points. In this case, the system is intersecting type.

Note 1. The equation to the limiting point (√d, 0, 0) is

⇒ \((x-\sqrt{d})^2+y^2+z^2=0 \text { i.e. } x^2+y^2+z^2-2 \sqrt{d} x+d=0\)

The equation to the limiting point (-√d, 0, 0) is

⇒ \((x+\sqrt{d})^2+y^2+z^2=0 \text { i.e. } x^2+y^2+z^2+2 \sqrt{d} x+d=0\)

2. If there is only one limiting point (0, 0, 0), its equation is x² + y² + z² = 0.

The Sphere Solved Problems

Example. 1. Find the limiting points of the coaxal system defined by spheres x² + y² + z² + 4x – 2y + 2z + 6 = 0 ….(1) and  x² + y² + z²+ 2x – 4y – 2z + 6 = 0       …..(2)

Solution:

Given

x² + y² + z² + 4x – 2y + 2z + 6 = 0 ….(1) and  x² + y² + z²+ 2x – 4y – 2z + 6 = 0       …..(2)

The R.P. of the sphere of the coaxal system is 2x + 2y + 4z = 0 i.e., x + y + 2z = 0

∴ The equation to a sphere of the coaxal system is

⇒ x² + y² + z² + 4x – 2y + 2z + 6 + λ(x + y + 2z) = 0

i.e. x² + y² + z² + (4+λ)x + (λ-2)y + (2λ+2)z + 6 = 0

∴ centre = \(\left(-\frac{4+\lambda}{2}, \frac{2-\lambda}{2},-\lambda-1\right)\) and

radius = \(\sqrt{\left[\frac{(4+\lambda)^2}{4}+\frac{(2-\lambda)^2}{4}+(\lambda+1)^2-6\right]}\)

For limiting points of the system, radius = 0.

∴ \(\frac{(4+\lambda)^2}{4}+\frac{(2-\lambda)^2}{4}+(\lambda+1)^2-6=0\) i.e., λ2 + 2λ = 0

i.e. λ = 0, -2

∴ Limiting points are (-2, 1, -1); (-1, 2, 1).

Geometric Interpretation Of Radical Planes And Orthogonal Spheres

Example.2. Find the equation of the sphere belonging to the coaxal system given by x² + y² + z² – 2ax – 2ay – 2az + 4a2 + λ(x + y – z) = 0 and x² + y² + z² – 4ax – 4ay + 4a2 = 0 and which cuts the sphere x² + y² + z² – 2ax = 0 orthogonally.

Solution:

Given

x² + y² + z² – 2ax – 2ay – 2az + 4a2 + λ(x + y – z) = 0 and x² + y² + z² – 4ax – 4ay + 4a2 = 0

x² + y² + z² – 2ax = 0

R.P. of the given spheres is 2ax + 2ay – 2az = 0 i.e., x + y – z =0

∴ The equation to a sphere of the coaxal system is

⇒ x² + y² + z² – 2ax – 2ay – 2az + 4a2 + λ(x + y – z) = 0

If the sphere intersects x² + y² + z² – 2ax – 2ay – 2az = 0 orthogonally, then

⇒ \(2\left(\frac{\lambda-2 a}{2}\right) \cdot a+2\left(\frac{\lambda-2 a}{2}\right) \cdot 0+2\left(\frac{-\lambda-2 a}{2}\right) \cdot 0=4 a^2\) i.e., λ=6a.

∴ Equation to the required sphere is x² + y² + z² – 4ax – 4ay – 8az + 4a2 = 0.

Solved Problems On Equations Of Spheres And Their Radical Planes

Example. 3. Prove that every sphere through the limiting points of a coaxal system intersects every sphere of that system orthogonally.

Solution: Let a coaxal system of spheres be x² + y² + z² + 2λx + d = 0 …..(1)

(d > 0)

∴ Limiting points of the system are (-√d, 0, 0), (√d, 0, 0)

Let a sphere through the limiting points be x² + y² + z² + 2ux + 2vy + 2wz + c = 0 ……(2)

∴ \(d-2 u \sqrt{d}+c=0, d+2 u \sqrt{d}+c=0\)

∴ Solving, u=0, c = -d.

∴ The equation to the system of spheres through the limiting points is x² + y² + z² + 2vy 2wz – d = 0 …..(3)

Since 2.λ.0 + 2.0.v + 2.0.w = d – d is true for any sphere of the system (1), every sphere of the system intersects every sphere of the system through the limiting points orthogonally.

The Cone

Theorems Related To Cones With Proofs And Examples

Definition. The surface generated by a straight line that which passes through a fixed point and intersecting a given curve or touches a given surface, is called a cone.

The fixed point is called the vertex and the given curve the guiding curve of the cone.

An individual straight line on the surface of a cone is called a generator.

Thus a cone is the set of lines called generators through a given point.

Another Definition. Let S be a set of points in space. It there exists a point V in S such that P ∈ S ⇒ \(\overleftrightarrow{V P}\) ⊂ S then S is called the cone and V is said to be the vertex of the cone.

⇒\(\overleftrightarrow{V P}\) is called a generator of a cone.

Note 1. If V is the vertex of the cone S and P is a point on ‘S’ the \(\overleftrightarrow{V P}\) is a generator.

2. If L is a generator of the cone S then every point of L lies on S.

Example. (1) The equation 2×2 + 3y2 – z2 = 0 represents a cone with vertex as origin.

(2) Intersection pairs of planes form a cone with every point on the common line as a vertex.

(3) A plane is a cone with every point on it as a vertex.

Theorem.1  If f(x, y, z) is a homogeneous polynomial of nth degree then the surface S represented by f(x, y, z) = 0 is a cone with a vertex at the origin.

Proof. Since f(x, y, z) is a homogeneous polynomial of degree n, for a real number λ.

f(λx, λy, λz) = λⁿ f(x, y, z).

⇒ λⁿf(x, y, z) = 0 ⇒ f(λx, λy, λz) = 0

⇒ every point on \(\overleftrightarrow{O P}\) lies on the surface S.

∴ P ∈ S ⇒ \(\overleftrightarrow{O P}\) ⊂ S.

Hence the homogeneous equation f(x, y, z) = 0, represents a cone with a vertex at the origin.

Corollary. The line \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) is (λl, λm, λn) where λ is a real number.

∴ \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) is a generator of the cone <=> f(l,m,n) = 0

<=> Any point on the generator ∈ cone <=> (λl, λm, λn) ∈ cone

<=> f(λl, λm, λn) = 0 <=> λⁿ f(l, m, n) = 0 <=> f(l, m, n) = 0

NOTE. If f(x, y, z) is a homogeneous polynomial of degree n then the cone f(x, y, z) = 0 is called a cone of nth degree.

example (1) 2x³ – y³ + 3x²z + 2z³ = 0 is a cone of third degree.

(2) x² + y² – z² = 0 is a cone of 2nd degree.

Cones of second degree are also called Quadric cones. In this chapter, we deal with quadric cones only.

It will be seen that the degree of equation of a cone whose generators intersect a given conic or touch a given sphere is of the second degree.

Quadric Cones With Vertex At The Origin

Theorem.2  The equation of a cone with vertex at the origin is a homogeneous equation.

Proof. Let the general equation of second degree

ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 represent cone S with vertex at (0, 0, 0).

Let P (x₁, y₁, z₁) be a point on the cone

∴ The equation of generator \(\overleftrightarrow{O P}\) is \(\frac{x}{x_1}=\frac{y}{y_1}=\frac{z}{z_1}\) (=λ)

Any point (λx₁, λy₁, λz₁) on \(\overleftrightarrow{O P}\) lies on the cone S.

<=> λ²(ax₁² + by₁² + cz₁² + 2fy₁z₁ + 2gz₁x₁ + 2hx₁y₁) + 2λ(ux₁ + vy₁ + wz₁) + d = 0

This is true for all real values of λ.

<=> ax₁² + by₁² + cz₁² + 2fy₁z₁ + 2gz₁x₁ + 2hx₁y₁ …..(1)

ux₁ + vy₁ + wz₁ …..(2)

d = 0                 …..(3)

The relation (3) is obvious as the origin lies on the cone.

If u, v, w are not all zero, equation(2)

⇒ (λx₁, λy₁, λz₁) lies on the plane ux + vy + wz = 0

Which is a contradiction. Thus we have u = v = w = 0, d = 0

Hence the equation to the cone S with vertex at the origin is given by the homogeneous equation ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 …..(1)

Conversely. Every homogeneous equation of the second degree represents a cone with its vertex at the origin.

Let the homogeneous equation of second degree be

S ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

Let P(x₁, y₁, z₁) be a point on the cone.

Then ax₁² + by₁² + cz₁² + 2fy₁z₁ + 2gz₁x₁ + 2hx₁y₁ …..(1)

Multiplying by a real number λ2 we have

aλ²x₁² + bλ²y₁² + cλ²z₁² + 2fλ²y₁z₁ + 2hλ²x₁y₁ = 0

⇒ (λx₁, λy₁, λz₁) lies on the surface.

Thus P lies on the surface ⇒ Every point on OP lies on it.

∴ The surface is generated by lines through O and hence, by definition is a cone with its vertex at O.

Note.1. Let △ = abc + 2fgh – af² – bg² – ch² = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\)

(1) If △ = 0, equation I represents a pair of planes, and S is called a degenerate cone.

(2) If △ ≠ 0, then the surface S is called a Quadric cone or a non-degenerate cone.

2. The equation ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 always represents a cone of second degree with its vertex at the origin.

3. As the above equation of the cone contains five arbitrary constants, we need five conditions to determine the cone.

4. The general equation of the cone with vertex at (α, β, γ) is

a(x−α)²+b(y−β)²+c(z−γ)²+2f(y−β)(z−γ)+2g(z−γ)(x−α)+2h(x−α)(γ−β)=0

This is a homogeneous equation in (x-α),(y-β), and (z-γ).

The Cone Solved Problems

Example. 1. Find the equation of the cone whose generators pass through the point (α, β, γ) and have their direction cosines satisfying the relation al² + bm² + cn² = 0.

Solution. Equation to the generator passing through (α, β, γ)

and having directions cosines (l, m, n) is \(\frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}=k \text { (say) }\)

⇒ \(l=\frac{x-\alpha}{k}, m=\frac{y-\beta}{k} \text { and } n=\frac{z-\gamma}{k}\)

But l, m, n satisfy al² + bm² + cn² = 0 <=> \(\frac{1}{k}\left[(x-\alpha)^2+(y-\beta)^2+(z-\gamma)^2\right]=0\)

Hence the required to the cone is \((x-\alpha)^2+(y-\beta)^2+(z-\gamma)^2=0\)

Homogeneous Cone With Vertex At The Origin Examples

Example. 2. Show that x = -y = -z is a generator of the cone 5yz + 8zx – 3xy = 0.

Solution. \(\frac{x}{1}=\frac{y}{-1}=\frac{z}{-1}\)

is a generator of the cone 5yz + 8zx – 3xy = 0 …..(1)

<=> 5(-1)(-1)+8(-1)(1)-3(1)(-1) ⇒ 5 – 8 + 3 = 0

Hence the given line is a generator of the cone (1)

Theorem.3 shows that the general equation of the cone of the second degree which passes through the co-ordinate axes is fyz + gzx + hxy = 0

Proof. The equation of the cone of the second degree is ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 …..(1)

X – axis is a generator of the cone.

⇒ the direction cosines (1, 0, 0) of the x-axis satisfies (1)

⇒ a = 0.

Similarly, the y-axis is a generator ⇒ b = 0.

Similarly, the z-axis is a generator ⇒ c = 0.

Hence the general equation of the cone containing the three axes is fyz + gzx + hxy = 0

The Cone Solved Problems

Example. 1. Show that a cone can be found so as to contain any two given sets of three mutually perpendicular concurrent lines as generators.

solution. Let one set of three mutually perpendicular concurrent lines be taken as co-ordinate axes.

∴ The general equation of the cone through the coordinate axes is fyz + gzx + hxy = 0 …..(1)

Let the other set of perpendicular lines be OA, OB, OC given by the equations

Let the other set of lines be OA, OB, OC given by the equations

⇒\(\frac{x}{l_1}=\frac{y}{m_1}=\frac{z}{n_1} ; \frac{x}{l_2}=\frac{y}{m_2}=\frac{z}{n_2} ; \frac{x}{l_3}=\frac{y}{m_3}=\frac{z}{n_3}\)

Then m₁n₁ + m₂n₂ + m₃n₃ = 0 …..(1)

n₁l₁ + n₂l₂ + n₃l₃ = 0                …..(2)

l₁m₁ + l₂m₂ + l₃n₃ = 0             …..(3)

OA, OB, OC are the generators of cone

⇒ fm₁n₁ + gn₁l₁ + hl₁m₁ = 0   …..(2)

fm₂n₂ + gn₂l₂ + hl₂m₂ = 0      …..(3)

Adding (2) and (3) we get f(m₁n₁ + m₂n₂) + g(n₁l₁ + n₂l₂) + h(l₁m₁ + l₂m₂) = 0

i.e. f(-m₃n₃) + g(-n₃l₃) = 0       …by(3)

i.e. fm₃n₃ + gn₃l₃ + hl₃m₃ = 0

⇒ the line OC with directions ratio (l₃, m₃, n₃) lies on the cone fyz + gzx + hxy = 0.

Hence cone (1) contains two sets of mutually perpendicular generators.

Step-By-Step Guide To Solving Cone Geometry Problems

Example. 2. Find the equation to the cone which passes through the three coordinate axes as well as the three lines \(\frac{1}{2} x=y=-z, x=\frac{1}{3} y=\frac{1}{5} z \text { and } \frac{1}{8} x=-\frac{1}{11} y=\frac{1}{5} z\).

Solution. The equation to the cone passing through the three coordinate axes can be taken in the form fyz + gzx + hxy = 0 …..(1)

The line \(\frac{x}{2}=\frac{y}{1}=\frac{z}{-1}\) lies on (1)

<=> f(1)(-1)+g(-1)(2)+h(2)(1) = 0 ⇒ f + 2g – 2h = 0       …..(2)

The line \(\frac{x}{1}=\frac{y}{3}=\frac{z}{5}\) lies on (1)

<=> f(3)(5) + g(5)(1) + h(1)(3) = 0 ⇒ 15f + 5g + 3h = 0  …..(3)

Solving (2) and (3) : \(\frac{f}{6+10}=\frac{g}{-30-3}=\frac{h}{5-30} \Rightarrow \frac{f}{16}=\frac{g}{-33}=\frac{h}{-25}\)

Hence the equation of the cone (1) is 16yz – 33zx – 25xy = 0

Clearly, the line with d.r’s (8, -11, 5) lies on it.

Example. 3. Find the equation of the cone which contains the three coordinate axes and the two lines through the origin with direction cosines (l₁, m₁, n₁) and (l₂, m₂, n₂)

Solution. The equation to the cone containing the coordinate axes can be taken as fyz + gzx + hxy = 0 …..(1)

The two lines with d.c.’s (l₁, m₁, n₁) and (l₂, m₂, n₂) lie on cone (1)

<=> fm₁n₁ + gn₁l₁ + hl₁m₁ = 0   …..(2)

and f(m₂n₂) + gn₂l₂ + hl₂m₂ = 0  …..(3)

Solving (2) and (3):

⇒ \(\frac{f}{l_1 l_2 m_2 n_1-l_1 l_2 m_1 n_2}=\frac{g}{m_1 m_2\left(n_2 l_1\right)-m_1 m_2 n_1 l_2}=\frac{h}{n_1 n_2 l_2 m_1-n_1 n_2 l_1 m_2}\)

⇒ \(\frac{f}{l_1 l_2\left(m_2 n_1-m_1 n_2\right)}=\frac{g}{m_1 m_2\left(n_2 l_0-n_1 l_2\right)}=\frac{h}{n_1 n_2\left(l_2 m_1-l_1 m_2\right)}\)

Substituting in (1) the required equation of the cone is

⇒ \(l_1 l_2\left(m_2 n_1-m_1 n_2\right) y z+m_1 m_2\left(n_2 l_1-n_1 l_2\right) z x+n_1 n_2\left(l_2 m_1-l_1 m_2\right) x y=0\)

⇒ \(\sum l_1 l_2\left(m_1 n_2-m_2 n_1\right) y z=0\)

The Cone Cone And A Plane Through Its Vertex.

Find the angle between the lines of intersection of the plane px + qy + rz = 0 and the cone F(x, y, z) ≡ ax² + by² + cz²+ 2fyz + 2gzx + 2hxy = 0

Solution. Let a line of intersection of the plane with the cone be \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\).

∴ The line lies in the given plane as well as on the cone <=> pl + qm + rn = 0

and al² + bm² + cn² + 2fmn + 2gnl + 2hlm = 0 …..(2)

Now substituting \(n=-\frac{p l+q m}{r}\) in (2) we have

⇒ \(a l^2+b m^2+c\left(\frac{p l+q m}{r}\right)^2+(2 f m+2 g l)\left(\frac{-p l+q m}{r}\right)+2 h l m=0\)

⇒ \(l^2\left(c p^2+a r^2-2 g r p\right)+2 l m\left(c p q+h r^2-g p r-f r p\right)+m^2\left(b r^2+c q^2-2 f q r\right)=0\)

⇒ \(\frac{l^2}{m^2}\left(c p^2+a r^2-2 g r p\right)+\frac{2 l}{m}\left(c p q+h r^2-g p r-f r p\right)+\left(b r^2+c q^2-2 f q r\right)=0\) …..(3)

This is a quadratic equation in \(\frac{l}{m}\) and shows that the plane cuts the cone in two times. If (l1, m1, n1) and (l2, m2, n2) are the d.c’s of the two lines then

⇒ \(\frac{l_1}{m_1}, \frac{l_2}{m_2}\) are the roots of (3)

⇒ \(\frac{l_1}{m_1} \cdot \frac{l_2}{m_2}=\frac{b r^2+c q^2-2 f q r}{c p^2+a r^2-2 g r p}\)

⇒ \(\frac{l_1 l_2}{b r^2+c q^2-2 f q r}=\frac{m_1 m_2}{c p^2+a r^2-2 g r p}=\frac{n_1 n_2}{a q^2+b p^2-2 h p q}\) by symmetry

each = \(\frac{l_1 l_2+m_1 m_2+n_1 n_2}{(b+c) p^2+(c+a) q^2+(a+b) r^2-2 f q r-2 g r p-2 h p q}\)

Also sum of the roots of (3) is \(\frac{l_1}{m_1}+\frac{l_2}{m_2}=-\frac{2\left(c p q+h r^2-g p r-f r p\right)}{c p^2+a r^2-2 g r p}\)

⇒ \(\frac{l_1 m_2+l_2 m_1}{-2\left(c p q+h r^2-g p r-f r p\right)}=\frac{m_1 m_2}{c p^2+a r^2-2 g r p}=\frac{l_1 l_2}{b r^2+c q^2-2 f q r}=\frac{n_1 n_2}{a q^2+b p^2-2 h p q}\)

each = \(\frac{\left[\left(l_1 m_2+l_2 m_1\right)^2-4 l_1 l_2 m_1 m_2\right]^{1 / 2}}{\left[4\left(c p q-g p r-f r p+h r^2\right)-4\left(b r^2+c q^2-2 f q r\right)\left(c p^2+a r^2-2 g r p\right)\right]^{1 / 2}}\)

⇒ \(\frac{l_1 m_2-l_2 m_1}{\pm 2 r \mathrm{D}} \text { where } \mathrm{D}^2=\left|\begin{array}{cccc}
a & h & g & p \\
h & b & f & q \\
g & f & c & r \\
p & q & r & 0
\end{array}\right|\)

By symmetry \(\frac{l_1 m_2-l_2 m_1}{\pm 2 r \mathrm{D}}=\frac{m_1 n_2-m_2 n_1}{\pm 2 p \mathrm{D}}=\frac{n_1 l_2-n_2 l_1}{\pm 2 q \mathrm{D}}=\frac{\sqrt{\sum\left(m_1 n_2-m_2 n_1\right)^2}}{\pm 2 \mathrm{D} \sqrt{p^2+q^2+r^2}}\)

Let θ be the angle between the lines then tanθ = =\(\frac{\sqrt{\sum\left(m_1 n_2-m_2 n_1\right)^2}}{l_1 l_2+m_1 m_2+n_1 n_2}\)

⇒ \(\tan \theta=\frac{2 \mathrm{D} \sqrt{\left(p^2+q^2+r^2\right)}}{(a+b+c)\left(p^2+q^2+r^2\right)-\mathrm{F}(p, q, r)}\)

Cor. Condition of perpendicularity.

If the lines of intersection of the plane px + qy + rz = 0 and the cone

ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 is a right angle then

θ = 90° ⇒ tanθ = tan90° = ∞’

⇒ (a + b + c)(p² + q² + r²)-F(p, q, r) = 0 which is the required condition.

The Cone Solved Problems

Example. 1. Find the equation of the lines of intersection of the plane 2x + y – z = 0 and the cone 4x² – y² + 3z² = 0

Solution.

Given

The plane 2x + y – z = 0 and the cone 4x² – y² + 3z² = 0

Let a line intersection of the plane with cone be \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) …..(1)

The line (1) lies on the plane and the cone <=> 2l + m – n = 0 …..(2)

and 4l² – m² + 3n² = 0 …..(3)

∴ n = 2l + m

Substituting the value of n in (3): 4l² – m² + 3(2l+m)² = 0 ⇒ 8l² + 6lm + m² = 0

⇒ (2l + m)(4l + m) = 0 ⇒ 2l + m = 0 …..(4)

and 4l + m = 0 …..(5)

(1) when 2l + m = 0, we have from (2): n = 0

⇒ \(\frac{l}{1}=\frac{m}{-2}=\frac{n}{0}\)

(2) Solving 2l + m – n = 0 …..(2)

and 4l + m + 0.n = 0 …..(3)

⇒ \(\frac{l}{0+1}=\frac{m}{-4-0}=\frac{n}{2-4} \Rightarrow \frac{l}{1}=\frac{m}{-4}=\frac{n}{-2}\)

Hence the two lines are \(\frac{x}{1}=\frac{y}{-2}=\frac{z}{0} \text { and } \frac{x}{1}=\frac{y}{-4}=\frac{z}{-2}\)

Solved Exercise Problems On Cone Geometry Step-By-Step

Example. 2. Find the equations of the lines of intersection of the plane 3x + 4y + z = 0 and the cone 15x² – 32y² – 7z² = 0= 0.

Solution.

Given

The plane 3x + 4y + z = 0 and the cone 15x² – 32y² – 7z² = 0= 0

Let a line of intersection be \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\)

The line belongs to 3x + 4y + z = 0 <=> 3l + 4m + n = 0 …..(1)

⇒ n = -(3l + 4m)

Also, the line lies on the given cone 15x² – 32y² – 7z² = 0

<=> 15l² – 32m² – 7n² = 0 = 0 …..(2)

Substituting the value of n in (2):  15l² – 32m² – 7(3l + 4m)² = 0

⇒ 2l² – 7lm + 6m² = 0 ⇒ (l+2m)(2l+3m)=0

⇒ l + 2m = 0 …..(3)

and 2l + 3m = 0 …..(4)

(1) solving 3l + 4m + n = 0 …..(1)

and l + 2m + 0.n = 0 …..(4)

(2) Again solving 3l + 4m + n = 0 …..(1)

and 2l + 3m + 0.n = 0 …..(4)

⇒ \(\frac{l}{0-3}=\frac{m}{2-0}=\frac{n}{9-8} \Rightarrow \frac{l}{-3}=\frac{m}{2}=\frac{n}{1}\)

∴ The two lines of intersection are \(\frac{x}{-2}=\frac{y}{1}=\frac{z}{-2} \text { and } \frac{x}{-3}=\frac{y}{2}=\frac{z}{1}\)

Example.3. Find the angle between the lines of intersection of the plane x – 3y + z = 0 and the cone x² – 5y² + z² = 0.

Solution.

Given cone x² – 5y² + z² = 0 and the plane x – 3y + z = 0.

Let \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) be one of the common lines of the cone and the plane.

∴ l² – 5m² + n² = 0 …..(1)

and \(l-3 m+n=0 \Rightarrow \frac{l+n}{3}=m\) …..(2)

Substituting (2) in (1) \(l^2-5\left(\frac{l+n}{3}\right)^2+n^2=0\)

⇒ 2l² – 5ln + 2n² = 0 ⇒ (2l – n)(l – 2n) = 0

⇒ 2l – n = 0 …..(3) and l – 2n = 0 …..(4)

Now solving (2) and (3) i.e. l – 3m + n = 0

2l + 0.m – n = 0

we have \(\frac{l}{3-0}=\frac{m}{2+1}=\frac{n}{0+6} \Rightarrow \frac{l}{1}=\frac{m}{1}=\frac{n}{2}\)

Again solving (2) and (4) i.e. l – 3m + n = 0, l + 0.m – 2n = 0

We have \(\frac{l}{6-0}=\frac{m}{1+2}=\frac{n}{0+3} \Rightarrow \frac{l}{2}=\frac{m}{1}=\frac{n}{1}\)

Hence the direction ratios of the two lines of intersection are (1, 1, 2) and (2, 1,1).

∴ The equations of the lies of intersection are \(\frac{x}{1}=\frac{y}{1}=\frac{z}{2} \quad \text { and } \quad \frac{x}{2}=\frac{y}{1}=\frac{z}{1}\)

If θ is the angle between the lines, the cosθ = \(\frac{1(2)+1(1)+2(1)}{\sqrt{(1+1+4)} \cdot \sqrt{(4+1+1)}}=\frac{5}{6}\)

⇒ \(\theta=\cos ^{-1} \frac{5}{6}\)

Properties Of Cones With Solved Examples And Exercises

Example.4. Show that the equation of the quadric cone which contains the three coordinate axes and the lines in which the plane x – 5y – 3z = 0 cuts the cone 7x² + 5y² – 3z² = 0 is yz + 10zx + 18xy = 0.

Solution. Let the plane x – 5y – 3z = 0 cut the cone 7x² + 5y² – 3z² = 0

along the line \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\)

⇒ l 5m – 3n = 0 …..(1)

and 7l² + 5m² – 3n² = 0 …..(2)

Eliminating l from (1) and (2): 7(5m + 3n)² + 5m² – 3n² = 0

⇒ 6m² + 7mn + 2n² = 0 ⇒ (2m + n)(3m + 2n) = 0

⇒ 2m + n = 0 …..(3) and 3m + 2n = 0 …..(4)

(1) Solving l – 5m – 3n = 0 …..(1)

and l.0 + 2m + n = 0 …..(3)

⇒ \(\frac{l}{-5+6}=\frac{m}{0-1}=\frac{n}{2-0} \Rightarrow \frac{l}{1}=\frac{m}{-1}=\frac{n}{2}\)

(2) Solving l – 5m – 3n = 0 …..(1)

and l.0 + 3m + 2n = 0 …..(4)

⇒ \(\frac{l}{-10+9}=\frac{m}{0-2}=\frac{n}{3-0} \Rightarrow \frac{l}{-1}=\frac{m}{-2}=\frac{n}{3}\)

∴ The two lines of intersection are \(\frac{x}{1}=\frac{y}{-1}=\frac{z}{2} \text { and } \frac{x}{1}=\frac{y}{2}=\frac{z}{-3}\)

Let the cone containing the coordinate axes be fyz + gzx + hxy = 0 …..(5)

<=> f(2)(-1) + g(2)(1) + h(1)(-1) = 0 ⇒ 2f – 2g + h = 0 …..(6)

and f(2)(-3) + g(-3)(1) + h(1)(2) = 0 ⇒ 6f + 3g – 2h = 0 …..(7)

Solving (6) and (7) : \(\frac{f}{4-3}=\frac{g}{6+4}=\frac{h}{6+12} \Rightarrow \frac{f}{1}=\frac{g}{10}=\frac{h}{18}\)

Hence the required cone is yz + 10zx + 18xy = 0

Example.5. Prove that the angle between the lines of intersection of the plane x + y + z = 0 with the cone ayz + bzx + cxy = 0 is π/3 if \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\).

Solution. Let a line of intersection be \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) …..(1)

(1) lies on the given plane and cone

<=> l + m + n = 0 …..(2) and amn + bnl + clm = 0 …..(3)

Eliminating n from (2) and (3)

-am(l + m)-bl(l + m) + clm = 0 ⇒ bl² + (a + b – c)lm + am² = 0

This is a quadratic in \(\frac{l}{m}\). Let the roots be \(\frac{l_1}{m_1}, \frac{l_2}{m_2}\)

∴ \(\frac{l_1}{m_1} \cdot \frac{l_2}{m_2}=\frac{a}{b} \Rightarrow \frac{l_1 l_2}{a}=\frac{m_1 m_2}{b}=\frac{n_1 n_2}{c}\) by symmetry

each =\(\frac{l_1 l_2+m_1 m_2+n_1 n_2}{a+b+c}=k\)

Again \(\frac{l_1}{m_1}+\frac{l_2}{m_2}=\frac{c-b-a}{b} \Rightarrow \frac{l_1 m_2+l_2 m_1}{c-b-a}=\frac{m_1 m_2}{b}=k \text { (say) }\)

Now (l₁m₂ – l₂m₁)² = (l₁m₂ + l₂m₁)² – 4l₁l₂m₁m₂

= \(k^2(c-b-a)^2-4(a k)(b k)=k^2\left[(c-b-a)^2-4 a b\right]\)

= \(k^2\left(a^2+b^2+c^2-2 a b-2 b c-2 c a\right)\)

Now tanθ = \(\frac{\sqrt{\sum\left(l_1 m_2-l_2 m_1\right)^2}}{l_1 l_2+m_1 m_2+n_1 n_2}=\frac{\sqrt{3 k^2\left(a^2+b^2+c^2-2 b c-2 c a-2 a b\right)}}{k(a+b+c)}\)

If \(\theta=\frac{\pi}{3}, \tan ^2 \frac{\pi}{3}=(\sqrt{3})^2=\frac{3\left(a^2+b^2+c^2-2 b c-2 a c-2 a b\right)}{(a+b+c)^2}\)

⇒ (a + b + c)² = a² + b² + c² – 2bc – 2ca – 2ab

⇒ 4(bc + ca + ab) = 0 ⇒ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Worked Examples Of Cones With The Vertex At The Origin

Example. 6. Prove that if the angle between the lines of intersection of the plane x + y + z = 0 and the cone ayz + bzx + cxy = 0 is \(\frac{\pi}{2}\) then a + b + c = 0.

Solution. Let a line of intersection be \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) …..(1)

(1) lies on the cone and the plane <=> l + m + n = 0 …..(2)

amn + bnl + clm = 0 …..(3)

Substituting n = -l – m in (3)

-am(l+m) – bl(l+m) + clm = 0 ⇒ -alm – am² – bl² – blm + clm = 0

⇒ bl2 + (a + b – c) lm + am2 = 0 ⇒ \(b\left(\frac{l}{m}\right)^2+(a+b-c) \frac{l}{m}+a=0\)

This is a quadratic in \(\frac{l}{m}\). Let the roots be \(\frac{l_1}{m_1}, \frac{l_2}{m_2}\)

∴\(\left(\frac{l_1}{m_1}\right)\left(\frac{l_1}{m_2}\right)=\frac{a}{b} \Rightarrow \frac{l_1 l_2}{a}=\frac{m_1 m_2}{b}\)

By symmetry: \(\frac{l_1 l_2}{a}=\frac{m_1 m_2}{b}=\frac{n_1 n_2}{c}=K \text { (say) }\)

the angle between the lines with dcs (l₁, m₁, n₁) and (l₂, m₂, n₂) is \(\frac{\pi}{2}\)

<=>l₁l₂ + m₁m₂ + n₁n₂ = 0

⇒ ka + kb + kc = 0

⇒ a + b + c = 0.

The Cone Cone With A Base Curve

Definition. Let S be the set of lies concurrent at V and C be a curve not containing V.

If P ∈ C ⇒ \(\overleftrightarrow{V P}\) ⊂ S then S is called the base curve or guiding curve. \(\overleftrightarrow{V P}\) is called a generator of the cone.

Theorem.4  The equation of a cone with vertex at (α, β, γ) ∉ XY plane and the guiding curve f(x,y)=0, z=0 is \((z-\gamma)^2 \cdot f\left(\alpha-\gamma \frac{X-\alpha}{z-\gamma}, \beta-\gamma \frac{y-\beta}{z-\gamma}\right)=0\)

Proof. Let the equation to a line through (α, β, γ) with directions ratios (l, m, n) be

⇒ \(\frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n} \quad(=r)\) …..(1)

A point on the line is (lr + α, mr + β, nr + γ).

Let the equation to the curve be f(x,y) = ax² + 2hxy + by² + 2gx + 2fy + c = 0, z = 0

The line passes through the conic.

<=> The point P(lr + α, mr + β, nr + γ) lies on the f(x,y) = 0 and on the plane z = 0.

⇒ \(z=0 \Rightarrow n r+\gamma \Rightarrow r=-\frac{\gamma}{n}\)

Hence the point P =\(\left(\alpha-\frac{l \gamma}{n}, \beta-\frac{m \gamma}{n}, 0\right)\) which will lie on the given conic,

<=> \(a\left(\alpha-\frac{l \gamma}{n}\right)^2+2 h\left(\alpha-\frac{l \gamma}{n}\right)\left(\beta-\frac{m \gamma}{n}\right)+b\left(\beta-\frac{m \gamma}{n}\right)^2+2 g\left(\alpha-\frac{l \gamma}{n}\right)+2 f\left(\beta-\frac{m \gamma}{n}\right)+c=0\) …..(3)

This is the condition for the line to intersect the conic.

Now eliminating l, m, n between (1) and (3)

⇒ \(a\left(\alpha-\frac{x-\alpha}{z-\gamma} \cdot \gamma\right)^2+2 h\left(\alpha-\frac{x-\alpha}{z-\gamma} \cdot \gamma\right)\left(\beta-\frac{y-\alpha}{z-\gamma} \cdot \gamma\right)+b\left(\beta-\frac{y-\beta}{z-\gamma} \cdot \gamma\right)^2\)

⇒ \(+2 g\left(\alpha-\frac{x-\alpha}{z-\gamma} \cdot \gamma\right)+2 f\left(\beta-\frac{y-\beta}{z-\gamma} \cdot \gamma\right)+c=0\)

⇒ \(a(\alpha z-x \gamma)^2+2 h(\alpha z-x \gamma)(\beta z-\gamma y)+b(\beta z-\gamma y)^2\)

⇒ \(+2 g(\alpha z-\gamma x)(z-\gamma)+2 f(\beta z-\gamma y)(z-\gamma)+c(z-\gamma)^2=0\)

⇒ \((z-\gamma)^2 \cdot f\left(\alpha-\frac{x-\alpha}{z-\gamma} \cdot \gamma, \beta-\frac{y-\beta}{z-\gamma} \cdot \gamma\right)=0\) which is the required equation of the conic.

Note. The guiding curve of the cone may be f(y,z) = 0, x = 0 or f(z,x) = 0, y = 0

The Cone Solved Problems

Example. 1. Find the equation of the cone whose vertex is the origin and whose base curve is x² + y² + z² + 2ux + d = 0 px + qy + rz = k.

Solution.

Given

x² + y² + z² + 2ux + d = 0 px + qy + rz = k

Let P(x₁, y₁, z₁) be a point on the cone.

∴ The equation to \(\overleftrightarrow{O P}\) is \(\frac{x}{x_1}=\frac{y}{y_1}=\frac{z}{z_1}(=r)\). A point on \(\overleftrightarrow{O P}\) is (λx₁, λy₁, λz₁).

⇒ \(\overleftrightarrow{O P}\) intersects the base curve C ⇒ (λx₁, λy₁, λz₁) ∈ C

<=> λ²(λx₁², λy₁², λz₁²) + 2uλx1 + d = 0 and λ(px₁ + qy₁ + rz₁) = k.

Eliminating λ from the above two relations.

We have k²(x₁² + y₁² + z₁²) + 2ukx₁(px₁ + qy₁ + rz₁) + d(px₁ + qy₁ + rz₁)² = 0

Hence the equation to the locus of p is the curve

k²(x² + y² + z²) + 2ukx (px + qy + rz) + d(px + qy + rz)² = 0

Alternative method:

Solution. The given curve is x² + y² + z² + 2ux + d = 0 …..(1)

px + qy + rz = k …..(2)

The required equation is the homogeneous equation of second degree satisfied by the points common to the two equations.

∴ p x+q y+r z=k. \(\quad \frac{p x+q y+r z}{k}=1\) …..(3)

Now homogeneuising the equation of (1) by (3) we have the homogeneous equation

⇒ \(x^2+y^2+z^2+2 u x\left(\frac{p x+q y+r z}{k}\right)+d\left(\frac{p x+q y+r z}{k}\right)^2=0\)

Thus the equation to the homogeneous cone is

⇒ \(k^2\left(x^2+y^2+z^2\right)+2 u k x(p x+q y+r z)+d(p x+q y+r z)^2=0\)

Example. 2. Find the equation of the cone whose vertex is (1, 1, 0) and whose guiding curve is y = 0, x² + z² = 4

Solution. Let the equation to the generator through the vertex (1, 1, 0) be

⇒ \(\frac{x-1}{l}=\frac{y-1}{m}=\frac{z}{n}(=r)\) …..(1)

A point on the generator is (lr + 1, mr + 1, nr)

If this point lies on the curve y = 0,  x² + z²  = 4, then

mr+1=0, (lr + 1)2 + (nr)2 = 4

i.e., \(r=-\frac{1}{m},(l r+1)^2+n^2 r^2=4\)

Eliminating r \(\left(1-\frac{l}{m}\right)^2+\frac{n^2}{m^2}=4 \Rightarrow(m-l)^2+n^2=4 m^2\) …..(2)

Eliminating l, m, n from (2) by using (1)

[(y – 1)+(x – 1)]² + z² = 4(y – 1)² ⇒ x² – 3y² + z² – 2xy + 8y – 4 = 0

which is the equation to the required cone.

Classification Of Cones And Their Equations With Solved Problems

Example. 3. Find the equation of the cone with vertex (5, 4, 3) and 3x² + 2y² = 6, y + z = 0 as base.

Solution. Let the equation to the generator be \(\frac{x-5}{l}=\frac{y-4}{m}=\frac{z-3}{n}=k\) …..(1)

Any point on (1) is (lk + 5, mk + 4, nk + 4)

This point lies on the base <=> 3(lk + 5)² + 2(mk + 4)² = 6 …..(2)

and mk + 4 + nk 3 = 0 ⇒ k(m + n) = -7 …..(3)

Substituting (3) in (2) : \(3\left(5-\frac{7 l}{m+n}\right)^2+2\left(4-\frac{7 m}{m+n}\right)^2=6\)

⇒ \(3(5 m+5 n-7 l)^2+2(4 m+4 n-7 m)^2=6(m+n)^2\)

⇒\(3[5(y-4)+5(z-3)-7(x-5)]^2+2[4(z-3)-3(y-4)]^2=6(y-4+z-3)^2\)

⇒ \(3(-7 x+5 y+5 z)^2+2(-3 y+4 z)^2=6(y+z-3)^2\)

⇒ \(147 x^2+87 y^2+101 z^2-210 x y+90 y z-210 x y-294=0\)

Example.4. Obtain the locus of the lines that pass through a point (α, β, γ) and through the points of the conic \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0\)

Solution. Let the equation line through (α, β, γ) be \(\frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}=k\) …..(1)

Any point on the line is (α+lk, β+mk, γ+nk)

The point lies on the conic \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0\)

<=> \(\frac{(\alpha+l k)^2}{a^2}+\frac{(\beta+m k)^2}{b^2}=1 \text { and } \gamma+n k=0 \Rightarrow k=-\frac{\gamma}{n}\)

Substituting the value k we have

⇒ \(\frac{1}{a^2}\left(\alpha-\frac{l \gamma}{n}\right)^2+\frac{1}{b^2}\left(\beta-\frac{m \gamma}{n}\right)^2=1 \Rightarrow \frac{(\alpha n-\gamma l)^2}{a^2 n^2}+\frac{(\beta n-m \gamma)^2}{b^2 n^2}=1\)

Eliminating l, m, n using (1)

⇒ \(\frac{1}{n^2 a^2 k^2}[\alpha(z-\gamma)-\gamma(x-\alpha)]^2+\frac{1}{n^2 b^2 k^2}[\beta(z-\gamma)-\gamma(y-\beta)]^2=1\)

⇒ \(\frac{(\alpha z-\gamma x)^2}{a^2}+\frac{(\beta z-\gamma y)^2}{b^2}=(z-\gamma)^2\)

Example. 5. Find the equation of the cone whose vertex is (1, 2, 3) and base y² = 4ax, z = 0.

Solution. Let a line through (1, 2, 3) be \(\frac{x-1}{l}=\frac{y-2}{m}=\frac{z-3}{n}=k\) …..(1)

Any point on the line is (1+lk, 2+mk, 3+nk)

It lies on the given conic <=> (2+mk)²= 4a(1+lk) and 3 + nk = 0 ⇒ k = -3/n

Eliminating k we have \(\left(2-\frac{3 m}{n}\right)^2=4 a\left(1-\frac{3 l}{n}\right) \Rightarrow(2 n-3 m)^2=4 a(n-3 l) n\)

using (1) we get [2(z-3)-3(y-2)]² = 4a[z-3-3(x-1)](z-3)

⇒ (2z-3y)² = 4a(y-3x)2(z-3)

Homogeneous Cone Equations With Proofs And Applications

Example. 6. The section of a cone whose vertex is P and guiding curve the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0\) by the plane z=0 is a rectangular hyperbola. Show that the locus of P is \(\frac{x^2}{a^2}+\frac{y^2+z^2}{b^2}=1\)

Solution.

Given

The section of a cone whose vertex is P and guiding curve the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, z=0\) by the plane z=0 is a rectangular hyperbola.

Let the point p(x₁, y₁, z₁)

Equation to a line through P be \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}=k\)

Any point on the lien is (x₁ + lk, y₁ + mk, z₁ + nk)

This lies on the conic

<=> \(\frac{\left(x_1+l k\right)^2}{a^2}+\frac{\left(y_1+m k\right)^2}{b^2}=1, z_1+n k=0 \Rightarrow k=-\frac{z_1}{n}\)

Eliminating k we have

⇒ \(\frac{1}{a^2}\left[x_1-\frac{l z_1}{n}\right]^2+\frac{1}{b^2}\left[y_1-\frac{m z_1}{n}\right]^2=1 \Rightarrow \frac{\left(n x_1-l z_1\right)^2}{a^2}+\frac{\left(n y_1-m z_1\right)^2}{b^2}=n^2\)

⇒\(\frac{1}{a^2}\left[x_1\left(z-z_1\right)-z_1\left(x-x_1\right)\right]^2+\frac{1}{b^2}\left[y_1\left(z-z_1\right)-z_1\left(y-y_1\right)\right]^2=\left(z-z_1\right)^2\)

⇒ \(\frac{\left(z x_1-x z_1\right)^2}{a^2}+\frac{\left(z y_1-y z_1\right)^2}{b^2}=\left(z-z_1\right)^2\)

Now this meets x = 0 in a curve \(\frac{z^2 x_1^2}{a^2}+\frac{\left(z y_1-y z_1\right)^2}{b^2}=\left(z-z_1\right)^2, x=0\)

This will be a rectangular hyperbola <=> coefficient of y²+ coefficient of z² = 0

⇒\(\frac{z_1^2}{b^2}+\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1\right)=0\).

Hence the locus P is \(\frac{x^2}{a^2}+\frac{y^2+z^2}{b^2}=1\)

Example.7. A cone has as base the circle  x² + y² + 2ax + 2by = 0, z = 0 and passes through the fixed point (0, 0, c). If the section of the cone by ZX plane is a rectangular hyperbola, Prove that the vertex lies on a fixed circle.

Solution.

Given

A cone has as base the circle  x² + y² + 2ax + 2by = 0, z = 0 and passes through the fixed point (0, 0, c). If the section of the cone by ZX plane is a rectangular hyperbola,

Let P(x₁, y₁, z₁) be the vertex of the cone and base curve

x² + y²  + 2ax + 2by = 0, z = 0

The line through P, \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\) …..(1)

meets the plane z = 0 at the point given by

⇒ \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n} \Rightarrow \text { at }\left(x_1-\frac{l z_1}{n}, y_1-\frac{m z_1}{n}, 0\right)\)

This point lies on the circle

<=> \(\left(x_1-\frac{l z_1}{n}\right)^2+\left(y_1-\frac{m z_1}{n}\right)+2 a\left(x_1-\frac{l z_1}{n}\right)+2 b\left(y_1-\frac{m z_1}{n}\right)=0\)

Eliminating (l, m, n) using (1)

⇒ \(\left(x_1-\frac{x-x_1}{z-z_1} \cdot z_1\right)^2+\left(y_1-\frac{y-y_1}{z-z_1} \cdot z_1\right)^2+2 a\left(x_1-\frac{x-x_1}{z-z_1} \cdot z_1\right)+2 b\left(y_1-\frac{y-y_1}{z-z_1} \cdot z_1\right)=0\)

⇒ \(\left(x_1 z-x z_1\right)^2+\left(y_1 z-y z_1\right)^2+2 a\left(x_1 z-x z_1\right)\left(z-z_1\right)+2 b\left(y_1 z-y z_1\right)\left(z-z_1\right)=0\) …..(1)

This cone passes through (0, 0, c)

<=> \(\left(x_1 c\right)^2+\left(y_1 c\right)^2+2 a\left(x_1 c\right)\left(c-z_1\right)+2 b\left(y_1 c\right)\left(c-z_1\right)=0\)

<=> \(\left(x_1^2+y_1^2\right) c+\left(2 a x_1+2 b y_1\right)\left(c-z_1\right)=0\) …..(2)

Again the section of the cone (1) by the plane y = 0 is

⇒ \(\left(x_1 z-x z_1\right)^2+\left(y_1 z\right)^2+2 a\left(x_1 z-x z_1\right)\left(z-z_1\right)+2 b y_1 z\left(z-z_1\right)=0\) …..(3)

(3) represents a rectangular hyperbola

<=> coefficient of z² + coefficient of y² = 0 ⇒ x₁² + y₁² + 2ax₁ + 2by₁ + z₁² = 0 …..(4)

Locus of P is given by (2) and (4) as \(\left(x^2+y^2\right) c+(2 ax+2 b y)(c-z)=0\) …..(5)

and x² + y² + z² + 2ax + 2by = 0 …..(6)

Multiplying (6) by c and subtracting from (5), we get

cz² + 2azx + 2byz = 0 ⇒ 2ax + 2by + 2cz = 0 …..(7)

Hence  x² + y² + z² + 2ax + 2by = 0 and 2ax + 2by + cz = together represent a circle.

The Cone Enveloping Cone

Definition. Let S be a surface and P be a point not on the surface. The set of tangent lines to the surface S and passing through P form a cone with vertex at P. This is called the enveloping cone or the tangent cone of the given surface.

Theorem.5. The enveloping cone of the sphere x² + y² + z² = a² with vertex at (x₁, y₁, z₁) is (xx₁ + yy₁ + zz₁ – a²)² = (x² + y² + z² – a²)(x₁² + y₁² + z₁² – a²).

Proof. Let S = x² + y² + z² – a² = 0

P(x₁, y₁, z₁) ∉ S = 0

⇒ x₁² + y₁² + z₁² – a² ≠ 0

Let Q(x, y, z) be a point on the enveloping cone C.

∴ \(\overleftrightarrow{\mathrm{PQ}}\) is a tangent line to the sphere S = 0.

Let R ∈ \(\overleftrightarrow{\mathrm{PQ}}\) and (R, P, Q) = λ : 1

∴ \(\mathrm{R}=\left[\frac{\lambda x+x_1}{\lambda+1}, \frac{\lambda y+y_1}{\lambda+1}, \frac{\lambda z+z_1}{\lambda+1}\right]\)

R ∈ S = 0

⇒ \(\left(\frac{\lambda x+x_1}{\lambda+1}\right)^2+\left(\frac{\lambda y+y_1}{\lambda+1}\right)^2+\left(\frac{\lambda z+z_1}{\lambda+1}\right)^2=a^2\)

⇒ \(\left(\lambda x+x_1\right)^2+\left(\lambda y+y_1\right)^2+\left(\lambda z+z_1\right)^2=a^2(\lambda+1)^2\)

⇒ \(\lambda^2\left(x^2+y^2+z^2-a^2\right)+2 \lambda\left(x x_1+y y_1+z z_1-a^2\right)+\left(x_1^2+y_1^2+z_1^2-a^2\right)=0\)

⇒ \(\lambda^2 S+2 \lambda S_1+S_{11}=0\)

If \(\overleftrightarrow{\mathrm{PQ}}\) is a tangent line to the sphere then the two roots of the equation (1) are equal

⇒ 4S₁² – 4SS₁₁ = 0 ⇒ s₁² = SS₁₁

Hence the equation to the enveloping cone C is s₁² = SS₁₁

i.e., (xx₁ + yy₁ + zz₁ – a²)² = (x² + y² + z² – a²)(x₁² + y₁² + z₁² – a²)

The Cone Right Circular Cone

Definition. A right circular cone is a surface generator by a line that passes through a fixed point and makes a constant angle with a fixed line through the fixed point.

Let S be a set of concurrent lines, concurrent at V. If there exists a line L passing through V such that for a line M, M ∈ S ⇒ (L, M) = θ the S is called a right circular cone with vertex at V.

The line L is called the axis is θ the semi-vertical angle of the cone.

Note. The section of a right circular cone by any plane perpendicular to the axis is a circle.

Theorem.6. The equation of a right circular cone with vertex at (α, β, γ), semi-vertical angle θ and axis having direction ratios (l, m, n) is [l(x – α) + m(y – β) + n(z – γ)]² = (l² + m² + n²)[(x – α)² + (y – β)² + (z – γ)²]cos²θ

Proof. Let V be the vertex and VL be the axis of the cone. V = (α, β, γ) and the direction ratios of the axis VL are (l, m, n).

Let P(x, y, z) be a point on the cone.

D.r’s of V.P are (x – α, y – β, z – γ)

Semi vertical angle θ = \((\overleftrightarrow{\mathrm{VL}}, \overleftrightarrow{\mathrm{VP}})\)

⇒ \(\cos \theta=\frac{l(x-\alpha)+m(y-\beta)+n(z-\gamma)}{\sqrt{\left[(x-\alpha)^2+(y-\beta)^2+(z-\gamma)^2\right]} \sqrt{\left(l^2+m^2+n^2\right)}}\)

Hence the equation of the right circular cone is

⇒ \(\left[(x-\alpha)^2+(y-\beta)^2+(z-\gamma)^2\right]\left(l^2+m^2+n^2\right) \cos ^2 \theta\)

⇒ \([l(x-\alpha)+m(y-\beta)+n(z-\gamma)]^2\)

Corollary 1. If the vertex be the origin then the equation of the cone becomes (lx + my + nz)² = (l² + m² + n²)(x² + y² + z²)cos²θ

Corollary 2. The equation of the right circular cone with vertex at (0, 0, 0) and whose axis is the z-axis and semi-vertical angle α is x² + y²  = z² tan²α

Proof. Since d.r’s of the z-axis are (0, 0, 1)

l = 0, m = 0, n = 1

∴ The equation to the right circular cone is (x² + y² + z² ) cos²α = z²

⇒ (x² + y²) = z² (sec²α – 1) ⇒ (x² + y²) = z² (tan²α)

The Cone Solved Problems

Example.1. Find the enveloping cone of the sphere x² + y² + z² + 2x – 2y = 2, with its vertex at (1, 1, 1).

Solution. Given Vertex = (1, 1, 1).

Equation to the given sphere is S = x² + y² + z² + 2x – 2y – 2 = 0

Now s₁ = x.1 + y.1 + z.1 + (x + 1) – (y + 1) – 2 = 0 = 2x + z – 2

S₁₁ = 1 + 1 + 1 + 2 – 2 – 2 = 1

∴ The equation to the enveloping cone is s₁² = SS₁₁

(2x + z – 2)² = (x² + y² + z² + 2x – 2y – 2)(1) ⇒ 3x² – y² + 4zx – 10x + 2y – 4z + 6 = 0

Example.2. Find the equation to the right circular cone whose vertex is P(2, -3, 5), axis PQ which makes equal angles with the axis and which passes through A(1, -2, 3).

Solution. The axis of the cone makes equal angles θ with the coordinate axes

∴ d.r’s of the axis are (cosθ, cosθ, cosθ) ⇒ d.r’s of the axis are (1, 1, 1)

Let α be the semi-vertical angle of the cone with vertex P(2, -3, 5)

∴ The equation to the required cone is

[(x – 2)² + (y + 3)² + (z – 5)²] (1 + 1 + 1)cos²α = [1.(x – 2) + 1.(y + 3) + 1(z – 5)]²

The point A(1, -2, 3) lies on the cone

<=> \(\left[(1-2)^2+(-2+3)^2+(3-5)^2\right] 3 \cos ^2 \alpha=[(1-2)+(-2+3)+(3-5)]^2\) <=> [/latex]\cos \alpha=\frac{\sqrt{2}}{3}[/latex]

∴ The equation to the required cone is

⇒ \(\left[(x-2)^2+(y+3)^2+(z-5)^2\right] \times \frac{2}{3}=[(x-2)+(y+3)+(z-5)]^2\)

Simplifying the equation x² + y² + z² + 6(yz + zx + xy) – 16x – 36y – 4z – 28 = 0

Example.3. Find the equation of the right circular cone with vertex at (2, 1, -3) and whose axis is parallel to OY and whose semi-vertical angle is 45°.

Solution. Axis is parallel OY ⇒ d.cs. of axis are (0, 1, 0)

Given semi vertical angle α = 45°, vertex = (2, 1, -3).

∴ Equation to the cone is [(x – 2)² + (y – 1)² + (z + 3)²] =(y – 1)² + (z + 3)²](0 + 1 + 0)cos²45

= [0.(x – 2) + 1.(y – 1) + 0.(z + 3)]²

⇒ \(\frac{1}{2}\left[(x-2)^2+(y-1)^2+(z+3)^2\right]=(y-1)^2 \Rightarrow(x-2)^2-(y-1)^2+(z+3)^2=0\)

⇒ x² – y² + z² – 4x + 2y + 6z + 12 = 0

Example.4. Find the equation of the right circular cone whose vertex is the origin, axis as the line x = t, y = 2t, z = 3t and whose semi-vertical angle is 60°.

Solution. Vertex (α, β, γ) = (0, 0, 0)

Equation to the axis \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=t\)

⇒ D.r’s of the axis (l, m, n) = (1, 2, 2)

Semi vertical angle = 60°

∴ Equation to the required cone is

[(x – 0)² + (y – 0)² + (z – 0)²][1² + 2² + 3²]cos²60° = [1.(x – 0) + 2.(y – 0) + 3.(z – 0)]²

⇒ \(\frac{14}{4}\left(x^2+y^2+z^2\right)=(x+2 y+3 z)^2\)

⇒ 7(x² + y² + z²) = 2(x² + 4y² + 9z² + 4xy + 12yz + 6zx)

⇒ 5x² – y² – 11z² – 24yz – 12zx – 8xy = 0

Example.5. Show that the plane z = 0 cuts the enveloping cone of the sphere x² + y² + z² = 11 which has its vertex at (2, 4, 1) in a rectangular hyperbola.

Solution. Let S ≡ x² + y² + z² – 11 = 0

Given point P = (2, 4, 1) = (x₁, y₁, z₁)

S₁ ≡ xx₁ + yy₁ + zz₁ – 11 ≡ x(2) + y(4) + z(1) – 11 = 2x + 4y + z – 11

S₁₁ ≡ x₁² + y₁² + z₁² – 11 ≡ (2)² + (4)² + (1)² – 11 = 10

∴ Equation to the enveloping cone is SS11 = S12

⇒ (x² + y² + z² – 11)(10) = (2x + 4y + z – 11)²

Where the plane z = 0 cuts the cone, then the equation to the conic is

10(x² + y² – 11) = (2x + 4y – 11)² ⇒ 6x² – 6y² – 16xy + 88y + 44x – 331

In this equation coefficient of x² + coefficient of y² = 6 – 6 = 0

⇒ The conic is a rectangular hyperbola

Example.6. Find the equation of the cone generated by rotating the line \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) about the line \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) as axis.

Solution. Given lines pass through the origin ⇒ vertex is the origin.

D.r’s of an axis are (a, b, c)

Semi-vertical angle = angle between the generator and the axis

⇒ \(\cos \theta=\frac{a l+b m+c n}{\sqrt{a^2+b^2+c^2} \sqrt{l^2+m^2+n^2}}\) …..(1)

∴ Equation to the cone is [(x – 0)² + (y – 0)² + (z – 0)²](a² + b² + c²).cos²θ

= [a(x – 0) + b(y – 0) + c(z – 0)]²

Using (1) we have ⇒ (x² + y² + z²)(al + bm + cn)² = (l² + m² + n²)(ax + by + cz)²

Example.7. If α is the semi-vertical angle of a right circular cone which passes through the lines OY, OZ, and x = y = z. Show that cosα = (9 – 4√3)^-1/2.

Solution. Let (l, m, n) be d.r’s of the axis of the cone

D.r.’s of OY are (0, 1, 0)

D.r.’s of OZ are (0, 0, 1)

α is the angle between the axis and OY

⇒ \(\cos \alpha=\frac{0 . l+1 \cdot m+0 . n}{\sqrt{0+1+0} \sqrt{l^2+m^2+n^2}}=\frac{m}{\sqrt{l^2+m^2+n^2}}\) …..(1)

Also α is the angle between the axis and OZ

⇒ \(\cos \alpha=\frac{0 . l+0 . m+1 . n}{\sqrt{0+0+1} \sqrt{l^2+m^2+n^2}}=\frac{n}{\sqrt{l^2+m^2+n^2}}\) …..(2)

From (1) and (2) m = n

Similarly angle between the axis and \(\frac{x}{1}=\frac{y}{1}=\frac{z}{1}\)is

⇒ \(\cos \alpha=\frac{1 . l+1 \cdot m+1 \cdot n}{\sqrt{1+1+1} \sqrt{l^2+m^2+n^2}}=\frac{l+m+n}{\sqrt{3\left(l^2+m^2+n^2\right)}}\) …..(3)

Equating (1) and (3) \(m=\frac{l+m+n}{\sqrt{3}}\)

⇒ \(l+m(1-\sqrt{3})+n=0 \Rightarrow l+m(1-\sqrt{3})+m=0\) (∵ m = u)

⇒ \(l=m(\sqrt{3}-2) \Rightarrow \frac{l}{\sqrt{3}-2}=\frac{m}{1}=\frac{n}{1}\)

∴ From (1) \(\cos \alpha=\frac{1}{\sqrt{(\sqrt{3}-2)^2+1+1}}=\frac{1}{\sqrt{9-4 \sqrt{3}}}=(9-4 \sqrt{3})^{-1 / 2}\)

Example.8. Lines are drawn through the origin with direction ratios (1, 2, 2), (2, 3, 6) and (3, 4, 12). Find the direction ratios of the axis of the right circular cone and hence show that its semi-vertical angle is cos^-1(1/√3). Also, find the equation of the cone.

Solution.

Given

Lines are drawn through the origin with direction ratios (1, 2, 2), (2, 3, 6) and (3, 4, 12).

Let (l, m, n) be the direction ratios of the axis of the right circular cone

Let α be the semi-vertical angle of cone

∴ Each given line is at α with the axis

(1) \(\cos \alpha=\frac{1 . l+2 \cdot m+2 \cdot n}{\sqrt{1+4+4} \sqrt{l^2+m^2+n^2}}\) …..(1)

(2) \(\cos \alpha=\frac{2 \cdot l+3 \cdot m+6 \cdot n}{\sqrt{4+9+36} \sqrt{l^2+m^2+n^2}}\) …..(2)

(3) \(\cos \alpha=\frac{3 \cdot l+4 \cdot m+12 \cdot n}{\sqrt{9+16+144} \sqrt{l^2+m^2+n^2}}\) …..(3)

From (1) and (2) : \(\frac{l+2 m+2 n}{3}=\frac{2 l+3 m+6 n}{7} \Rightarrow l+5 m-4 n=0\) …..1

From (1) and (3): \(\frac{1}{3}(l+2 m+2 n)=\frac{1}{13}(3 l+4 m+12 n) \Rightarrow 2 l+7 m-5 n=0\) …..2

Solving 1 and 2: \(\frac{l}{-25+28}=\frac{m}{-8+5}=\frac{n}{7-10} \Rightarrow \frac{l}{3}=\frac{m}{-3}=\frac{n}{-3} \Rightarrow \frac{l}{1}=\frac{m}{-1}=\frac{n}{-1}\)

∴ Direction cosines of the axis are \(\left(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)\)

∴ The semi-vertical angle is given by

From (1) \(\cos \alpha=\frac{1(1)+2(-1)+2(-1)}{\sqrt{1+4+4} \sqrt{1+1+1}}=\frac{1}{\sqrt{3}} \Rightarrow \alpha=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)

Equation the cone is \(\cos \alpha=\frac{1}{\sqrt{3}}=\frac{1(x-0)-1(y-0)-1(z-0)}{\sqrt{1+1+1} \sqrt{x^2+y^2+z^2}}\)

⇒ (x – y – z)² = x² + y² + z² ⇒ yz – zx – xy = 0

Example.9. find the equation of the cone formed by rotating the line 2x + 3y = 6, z = 0 about the y – axis.

Solution. The direction cosines of the axis are (0, 1, 0)

Given equation to the generator is 2x + 3y = 6, z = 0

2x = -3(y – 2), z = 0

⇒ \(\frac{x}{3}=\frac{y-2}{-2}=\frac{z}{0}\) …..(1)

Also Y-axis meets the line 2x + 3y = 6, z = 0 at (0, 2, 0)

⇒ vertex of the plane = (0, 2, 0)

∴ Semi-vertical angle = Angle between the line (1) and Y – axis.

⇒ \(\cos \alpha=\frac{0.3+1(-2)+0.0}{\sqrt{0+1+0} \sqrt{9+4+0}}=\frac{-2}{\sqrt{13}}\)

∴ The equation to the right circular cone with vertex (0, 2, 0) and axis d.r.’s (0, 1, 0) is

[0(x – 0) + 1(y – 2) + 0.z]² (0 + 1 + 0)cos²α = (0.x + 1(y – 2) + 0.z)²

⇒ \([x+(y-2)+z]^2 \frac{4}{13}=(y-2)^2 \Rightarrow 4 x^2-9(y-2)^2+4 z^2=0\)

The Cone Notation

Let S represent the second-degree general equation in x, y, z. The following notation is used in this chapter.

i.e. S ≡ ax² + by² + cz²+ 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d

E = E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy

U = ax + hy + gz + u; V = hx + by + fz + v;

W = gx + fy + cz + w; D = ux + vy + wz + d and

U₁ = ax₁ + hy₁ + gz₁ + u; V₁ = hx₁ + by₁ + fz₁ + v;

W₁ = gx₁ + fy₁ + cz₁ + w; D₁ = ux₁ + vy₁ + wz₁ + d

Then S₁ = axx₁ + byy₁ + czz₁ + f(yz₁ + y₁z) + g(zx₁ + z₁x) + h(xy₁ + x₁y) + u(x + x₁) + v(y + y₁) + w(z + z₁) + d

= (ax₁ + hy₁ + gz₁ + u)x + (hx₁ + by₁ + fz₁ + v)y + (gx₁ + fy₁ + cz₁ + w) + ux₁ + vy₁ + wz₁ + d = U₁x + V₁y + W₁z + D₁

S₁₁ = U₁x₁ + V₁y₁ + W₁z₁ + D₁

Theorem.7. If (x₁, y₁, z₁) is the vertex of the cone S = 0 then U₁ = V₁ = W₁ = D₁ = 0

Proof. Let the equation to the cone be

S = ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0

Given vertex of the cone, P = (x₁, y₁, z₁)

Shifting the origin to the point P the new equation of the cone referred to vertex P as the new origin is

a(x + x₁)² + b(y + y₁)² + c(z + z₁)² + 2f(y + y₁)(z + z₁) + 2g(x + x₁)(z + z₁) + 2h(x + x₁)(y + y₁) + 2u(x + x₁) + 2v(y + y₁) + d = 0

⇒ ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2x(ax₁ + hy₁ + gz₁ + u) + 2y(hx₁ + by₁ + fz₁ + v) + 2z(gx₁ + fy₁ + cz₁ + w) + ax₁² + by₁² + cz₁² + 2fy₁z₁ + 2gz₁x₁ + 2hx₁y₁ + 2ux₁ + 2vy₁ + 2wz₁ + d = 0.

⇒ E(x, y, z) +2U₁x + 2V₁y + 2W₁z + S₁₁ = 0

This must be a homogeneous equation

⇒ U₁ = 0, V₁ = 0, W₁ = 0 and S₁₁ = 0.

But S₁₁ = U₁x₁ + V₁y₁ + W₁z₁ + D₁ = 0 ⇒ D₁ = 0 ⇒ U₁ = V₁ = W₁ = D₁ = 0

Corollary 1. If the equation S = 0 represents a cone then the condition is

⇒ \(\left|\begin{array}{llll}
a & h & g & u \\
h & b & f & v \\
g & f & c & w \\
u & v & w & d
\end{array}\right|=0\)

Proof. Eliminating x₁, y₁, z₁ in the equations

U₁ = ax₁ + hy₁ + gz₁ + u = 0; V₁ = hx₁ + by₁ + fz₁ + v = 0;

W₁ =gx₁ + fy₁ + cz₁ + w= 0

D₁ = ux₁ + vy₁ + wz₁ + d = 0

We get \(\left|\begin{array}{llll}
a & h & g & u \\
h & b & f & v \\
g & f & c & w \\
u & v & w & d
\end{array}\right|=0\)

This is the required condition that the equation S = 0 represents a cone.

Corollary 2. The vertex ( x₁, y₁, z₁ ) satisfies the equations

U ≡ ax + hy + bz + u = 0 …..(1) V ≡ hx + by + fz + v = 0 …..(2)

W ≡ gx + fy + cz + w = 0 …..(3) D ≡ ux + vy + wz + d = 0 …..(4)

Thus the vertex is obtained by solving any three of the above four equations

Note. Consider the homogeneous polynomial

S(x, y, z, t) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2uxt + 2vyt + 2wzt + dt2

Now \(\frac{\partial \mathrm{S}}{\partial x}=2(a x+h y+g z+u t)\)

⇒ \(\frac{\partial \mathrm{S}}{\partial y}=2(h x+b y+f z+v t) ; \quad \frac{\partial \mathrm{S}}{\partial z}=2(g x+f y+c z+w t) ; \quad \frac{\partial \mathrm{S}}{\partial t}=2(u x+v y+w z+d t)\)

Now equating \(\frac{\partial S}{\partial x}, \frac{\partial S}{\partial y}, \frac{\partial S}{\partial z}, \frac{\partial S}{\partial t}\) each to zero and putting t = 1, we get

U = V = W = D = 0.

The Cone Solved Problems

Example.1. If ax² + by² + cz² + 2ux + 2vy + 2wz + d = 0 represents a cone prove that \(\frac{u^2}{a}+\frac{v^2}{b}+\frac{w^2}{c}=d\).

Solution. Let ( x₁, y₁, z₁ ) be the vertex of the given cone.

The given equation represents a cone if

⇒ \(\mathrm{U}_1=0 \Rightarrow a x_1+u=0 \quad \Rightarrow x_1=\frac{-u}{a}\);

⇒ \(\mathrm{V}_{\mathrm{I}}=0 \quad \Rightarrow b y_1+v=0 \Rightarrow y_1=\frac{-v}{b}\)

⇒ \(\mathrm{W}_1=0 \Rightarrow c z_1+w=0 \Rightarrow z_1=\frac{-w}{c}\) and

D₁ = 0 ⇒ ux₁ + vy₁ + wz₁ + d = 0

Substituting in D₁ = 0 the values of x_{1 +}  y_{1 +}  z₁ we get

⇒ \(u\left(\frac{-u}{a}\right)+v\left(\frac{-v}{b}\right)+w\left(\frac{-w}{c}\right)+d=0\)

⇒ \(\frac{u^2}{a}+\frac{v^2}{b}+\frac{w^2}{c}=d\)

Example.2. Find the vertex of the cone 7x² + 2y² + 2z² – 10zx + 10xy + 26x – 2y + 2z – 17 = 0

Solution. Consider the homogeneous equation

S(x, y, z,t) = 7x² + 2y² + 2z² – 10zx + 10xy + 26xt – 2yt + 2zt – 17t2 = 0

∴ \(\frac{\partial \mathrm{S}}{\partial x}=14 x-10 z+10 y+26 t=14 x+10 y-10 z+26\) (∵ t = 1)

⇒ \(\frac{\partial S}{\partial y}=4 y+10 x-2 t=10 x+4 y-2\)

⇒ \(\frac{\partial \mathrm{S}}{\partial z}=4 z-10 x+2 t=-10 x+4 z+2\);

⇒ \(\frac{\partial \mathrm{S}}{\partial t}=26 x-2 y+2 z-34 t=26 x-2 y+2 z-34\)

Coordinates of vertex satisfy the equations

14x + 10y – 10z + 26 = 0 …..(1) 10x + 4y – 2 = 0 …..(2)

-10x + 4z + 2 =0 …..(3) 26x – 2y + 2z – 34 = 0 …..(4)

Solving (1), (2), and (3) we get x = 1, y = -2, z = 2

Substituting (1, -2, 2) in (4) 26 + 4 + 4 + – 34 = 0

Hence the vertex of the cone is (1, -2, 2)

Example.3. Show that the equation 2y² – 8yz – 4zx – 8xy + 6x – 4y – 2z + 5 = 0 represents a cone whose vertex is \(\left(-\frac{7}{6}, \frac{1}{3}, \frac{5}{6}\right)\)

Solution. Making the given equation homogeneous, we get

S(x, y, z, t) = 2y² – 8yz – 4zx – 8xy + 6xt – 4yt – 2zt + 5t2 = 0

⇒ \(\frac{\partial \mathrm{S}}{\partial x}=-4 z-8 y+6 t ; \quad \frac{\partial \mathrm{S}}{\partial y}=4 y-8 z-8 x-4 t\)

⇒ \(\frac{\partial \mathrm{S}}{\partial z}=-8 y-4 x-2 t ; \quad \frac{\partial \mathrm{S}}{\partial t}=6 x-4 y-2 z+10 t\)

Equating t = 1 coordinates of the vertex satisfy the equations

4y + 2z – 3 = 0 …..(1) 2x – y + 2z + 1 = 0 …..(2)

2x + 4y + 1 = 0 …..(3) 3x – 2y – z + 5 = 0 …..(4)

Solving (1),(2) and (3) we get \(x=-\frac{7}{6}, y=\frac{1}{3}, z=\frac{5}{6}\)

Substituting in (4): \(3\left(-\frac{7}{6}\right)-2\left(\frac{1}{3}\right)-\frac{5}{6}+5=0 \Rightarrow-21-4-5+30=0\)

Hence the vertex of the cone is \(\left(-\frac{7}{6}, \frac{1}{3}, \frac{5}{6}\right)\)

Theorem.8. The cone E(x, y, z) = 0 will have three mutually perpendicular generators <=> a + b + c = 0.

Proof. Given the equation of the cone

E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

Let \(\frac{x}{p}=\frac{y}{q}=\frac{z}{r}\) …..(1) be a generator of the cone,

∴ E(p, q, r) = 0 ⇒ ap² + bq² + cr² + 2fqr + 2grp + 2hpq = 0 …..(2)

The equation to the plane ⊥er to (1) and passing through the vertex is px + qy + rz = 0 …..(3)

Let this plane intersect the cone along two real generators and (l, m, n) be the d.c’s of one of the generators.

∴ al² + bm² + cn² + 2fmn + 2gnl + 2hlm = 0 …..(4)

pl + qm + rn = 0 …..(5)

Eliminating n between (4) and (5), we get

⇒ \(l^2\left(a r^2+c p^2-2 g r p\right)+2 l m\left(c p q+h r^2-g q r-f r p\right)+m^2\left(b r^2+c q^2-2 f q r\right)=0\)

⇒ \(\frac{l^2}{m^2}\left(a r^2+c p^2-2 g r p\right)+2 \frac{l}{m}\left(c p q+h r^2-g q r-f r p\right)+\left(b r^2+c q^2-2 f q r\right)=0\) …..(6)

If (l₁, m₁, n₁) and (l₂, m₂, n₂) are the direction cosines of the two generators of intersection then \(\frac{l_1}{m_1}, \frac{l_2}{m_2}\) are the roots of (6).

∴ \(\frac{l_1 l_2}{m_1 m_2}=\frac{b r^2+c q^2-2 f q r}{a r^2+c p^2-2 g r p}\)

⇒ \(\frac{l_1 l_2}{b r^2+c q^2-2 f g r}=\frac{m_1 m_2}{a r^2+c p^2-2 g r p}=\frac{n_1 n_2}{a q^2+b p^2-2 h p q}=k\), by symmetry.

∴ l₁l₂ + m₁m₂ + n₁n₂ = k[a(q² + r²) + b(r² + p²) + c(p² + q²) – 2fqr – 2grp – 2hpq]

= k(a + b + c)(p² + q² + r²) …..(2)

The two generators of intersection of the plane (3) with the cone are at right angles.

<=> l₁l₂ + m₁m₂ + n₁n₂ = 0

<=> a + b + c = 0 [(∵ (p, q, r) ≠ (0, 0, 0)

Since plane (3) is perpendicular to generator (1), the two generators of intersection of the plane (3) with the cone are perpendicular to generator (1).

∴ These three generators are mutually perpendicular

<=> Two generators of intersection are perpendicular <=> a + b + c = 0.

Note. 1. Above condition is satisfied whatever is the direction of the generator. From this, we get if three mutually perpendicular lines are generators to the cone, then a + b + c = 0

Note. 2. Let F(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 be a cone.

Shifting the origin to the vertex the transformed equation is

E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0.

the cone F(x, y, z) = 0, has three mutually perpendicular generators

<=> E(x, y, z) = 0 has three mutually perpendicular generators <=> a + b + c = 0

The Cone Solved Problems

Example.1. Show that the two lines of intersection of the plane ax + by + cz = 0 with the cone yz + zx + xy = 0 will be perpendicular if \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\).

Solution. Given cone is yz + zx + xy = 0.

In this equation Coefficient of x² + Coefficient of y² + Coefficient of z² = 0

∴ The cone contains sets of three mutually perpendicular generators.

The plane ax + by + cz = 0 cuts the cone in perpendicular generators if it’s a normal line through the vertex (0, 0, 0).

i.e., \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) is a generator of the cone.

⇒ bc + ca + ab = 0 ⇒ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Example.2. If the line \(x=\frac{1}{2} y=z\) represents one of the three mutually perpendicular generators of the cone 11yz + 6zx – 14xy = 0, find the equations of the other two.

Solution. The given cone is 11yz + 6zx – 14xy = 0

The plane through the vertex of the cone and perpendicular to the generator.

⇒ \(\frac{x}{1}=\frac{y}{2}=\frac{z}{1}\) …..(1) is x + 2y + z = 0 the other two generators perpendicular to (1) are the lines of intersection of 11yz + 6zx – 14xy = 0 and x + 2y + z = 0.

Let l, m, n be the direction ratios of one of the common lines.

Then 11mn + 6nl – 14lm = 0 …..(2)

and l + 2m + n = 0 ⇒ n = -1-2m

Substituting in (2) 11m(-l-2m) + 6l(-l -2m)-14lm = 0

⇒ 6l² + 37lm + 22m² = 0 ⇒ (2l + 11m)(3l + 2m) = 0

⇒ 2l + 11m = 0 or 3l + 2m = 0

(1) solving l + 2m + n = 0
2l + 11m + 0.n = 0

we get \(\frac{l}{-11}=\frac{m}{2}=\frac{n}{7}\)

(2) solving 1 + 2m + n = 0
3l + 2m + 0.n = 0

we get \(\frac{1}{-2}=\frac{m}{3}=\frac{n}{-4}\)

∴ the other two perpendicular generators are \(\frac{x}{-11}=\frac{y}{2}=\frac{z}{7} \text { and } \frac{x}{2}=\frac{y}{-3}=\frac{z}{4}\)

Example.3. Show that if a right circular cone has sets of three mutually perpendicular generators, its semi vertical angle must be tan^-1√2.

Solution. Let the origin be the vertex, l, m, n be direction cosines of the axis of the cone and α be its semi-vertical angle

Then the equation to cone is (lx + my + nz)² = (l² + m² + n²)(x² + y² + z²)cos²α

∵ the cone contains three mutually perpendicular generators, then

i.e., Coefficient of x² + Coefficient of y² + Coefficient of z² = 0.

Coefficient of x² = l²-(l² + m² + n²)cos²α

Coefficient of y² = m²-(l² + m² + n²)cos²α

Coefficient of z² = n²-(l² + m² + n²)cos²α

Adding, we have by (1) (l² + m² + n²)-3(l² + m² + n²)cos²α = 0

⇒ 1 – 3cos²α = 0 ⇒ tan²α = 2 ⇒ tanα = √2 ⇒ α = tan^-1 √2

Example.4. Show that cone whose vertex is the origin and which passes through the curve of intersection of the surface 2x² – y² + 2z² = 3d² any plane a distance d, from the origin has three mutually perpendicular generators.

Solution. The equation to any plane at a distance d from the origin is lx + my + nz = d …..(1)

where l, m, n are the actual d.c’s of normal to the plane.

Homogenizing the equation of the sphere with that of the plane, we have

⇒ \(2 x^2-y^2+2 z^2=3 d^2\left(\frac{l x+m y+n z}{d}\right)^2\)

Now Coefficient of x² + Coefficient of y² + Coefficient of z²

= (2 – 3l²) – l – 3m² + (2 – 3n²) = 3 – 3(l² + m² + n²) = 3 – 3(1) = 0

Hence plane (1) cuts the cone into three mutually perpendicular generators.

Example.5. If the plane 2x – y + cz = 0 cuts the cone yz + zx + xy = 0 in perpendicular lines find c.

Solution. Given cone yz + zx + xy = 0 …..(1)

contains sets of three mutually perpendicular generators.

2x – y + cz = 0 cuts (1) in perpendicular lines

⇒ the normal of the plane lies on it.

⇒ (2, -1, c) must satisfy the cone equation

⇒ (-1)(c) + c(2) + (2)(-1) = 0 ⇒ c = 2

Example.6. Find the locus of the point from which three mutually perpendicular lines can be drawn to intersection the central conic ax² + by² = 1, z = 0.

Solution. Let the point P be (x₁, y₁, z₁)

Any line through P is \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}=k\) …..(1)

Any point on the line is (x1 + lk, y1 + mk, z1 + nk)

the point lies on the base curve ax² + by² = 1, z = 0

<=> a(x₁ + lk)² + b(y₁ + mk)² = 1, z₁ + nk = 0

Eliminating k, we have \(a\left(x_1-\frac{l z_1}{n}\right)^2+b\left(y_1-\frac{m z_1}{n}\right)^2=1\)

⇒ a(nx₁ – lz₁)² + b(ny₁ – mz₁) = n²

using (1), to the cone is

a[x₁(z – z₁) – z₁(x – x₁)]² + b[y₁(z – z₁) – z₁(y – y₁)]² = (z – z₁)²

This contains three mutually perpendicular generators

<=> Coefficient of x² + coefficient of y²+ Coefficient of z² = 0 ⇒ az₁² + bz₁² + ax₁² + by₁² – 1 = 0.

∴ Locus of P is a(x² + z²) + b(y² + z²) = 1

The Cone Intersection Of A Line With A Cone

Let the equation to the cone S be

S(x, y, z) ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy + 2ux + 2vu + 2wz + d = 0

Let the equation to a line be \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}(=r)\)

Let P be a point on this line

∴ p = (lr + x₁, mr + y₁, nr + z₁) = 0

∴ P ∈ s <=> S(lr + x₁, mr + y₁, nr + z₁) = 0

<=> \(a\left(l r+x_1\right)^2+b\left(m r+y_1\right)^2+c\left(n r+z_1\right)^2+2 f\left(m r+y_1\right)\left(n r+z_1\right)+2 g\left(n r+z_1\right)\left(l r+x_1\right)\)

⇒ \(+2 h\left(l r+x_1\right)\left(m r+y_1\right)+2 u\left(l r+x_1\right)+2 v\left(m r+y_1\right)+2 w\left(n r+z_1\right)+d=0\).

<=> \(r^2\left(a l^2+b m^2+c n^2+2 f m n+2 g m l+2 h l m\right)+2 r\left[l\left(a x_1+h y_1+g z_1+u\right)\right.\)

⇒ \(\left.+m\left(h x_1+b y_1+f z_1+v\right)+n\left(g x_1+f y_1+c z_1+w\right)\right]+\mathrm{S}\left(x_1, y_1, z_1\right)=0\)

<=> \(r^2 \mathrm{E}(l, m, n)+2 r\left[l \mathrm{U}_1+m \mathrm{~V}_1+n \mathrm{~W}_1\right]+\mathrm{S}_{11}=0\)

(1) This will be a quadratic equation in r <=> E(l, m, n) ≠ 0

The equation will have two real and distinctive roots.

<=> (lU₁ + mV₁ + nW₁)² – E(l, m, n) S₁₁ > 0

Then there will be two real points of the line common with the cone. The line segment joining the two point is called the chord of the cone.

(2) If E(l, m, n) ≠ 0 and (lU₁ + mV₁ + nW₁)² – E(l, m, n) S₁₁ > 0 then there are no common points.

(3) If E(l, m, n) ≠ 0 and (lU₁ + mV₁ + nW₁)²   =  E(l, m, n) S₁₁ then the two roots of the equation are real and equal. Hence the line meets the curve in two coincident points. Then the line is called the tangent line at that common point.

(4) If E(l, m, n) = lU₁ + mV₁ + nW₁ = S₁₁ = 0, then the line becomes a generator of the cone.

The Cone Tangent Plane

Definition. Let S = 0, be the cone and L be a tangent line to the cone at P on it. The locus of the line L is called the tangent plane to the cone at P.

Theorem.9. If P(x₁, y₁, z₁) is a point on the cone S = 0, then the equation of the tangent plane to the cone at P is S₁ = 0.

Proof. Given equation to the cone S = S(x, y, z) = 0

Let the equation to the line passing through P(x₁, y₁, z₁) be

⇒ \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}(=r)\) …..(1)

P(x₁, y₁, z₁) ∈ S = 0 ⇒ S₁₁ = 0

The point (lr + x₁, mr + y₁, nr + z₁) of the line (1) lies on S = 0

<=> S(lr + x₁, mr + y₁, nr + z₁) = 0

<=> r²E(l, m, n)⁰ + 2r[lU₁ + mV₁ + nW₁] + S₁₁ < 0

The line is a tangent line to the cone

<=> (lU₁ + mV₁ + nW₁)² – E(l, m, n).S11 = 0

<=> lU₁ + mV₁ + nW₁ = 0 …..(2)

Eliminating l, m, n in (1) and (2), the locus of the tangent line is

(x – x₁)U₁ + (y – y₁)V₁ + (z – z₁)W₁ = 0

i.e. U₁x + V₁y + W₁z = U₁x₁ + V₁y₁ + W₁z₁

i.e. S₁ = S₁₁ i.e. S₁ = 0 [∵ S₁₁ = 0]

∴ The equation to the tangent plane at P(x₁, y₁, z₁) to the cone S = 0 is S₁ = 0

Corollary. If the equation of the cone

E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

then the equation to the tangent plane at (x₁, y₁, z₁) on the cone is U₁x + V₁y + W₁z = 0

i.e. x(ax₁ + hy₁ + gz₁ ) + y(hx₁ + by₁ + fz₁) + z[gx₁ + fy₁ + cz₁ ] = 0

i.e. axx₁ + byy₁ + czz₁ + f(y₁z + yz₁) + g(z₁x + zx₁) + h(x₁y + xy₁) = 0

Note 1. The tangent plane at a point P to the cone is also the tangent plane at every point on the generator.

2. The tangent plane at point P to the cone contains the generator through P.

3. The equation to the normal line to the tangent plane at P(x1, y1, z1) is \(\frac{x-x_1}{\mathrm{U}_1}=\frac{y-y_1}{\mathrm{~V}_1}=\frac{z-z_1}{\mathrm{~W}_1}\)

Theorem.10. The necessary and sufficient condition for the plane π = lx + my + nz = 0 to be a tangent plane to the cone E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 is \(\rho=\left|\begin{array}{llll}
a & h & g & l \\
h & b & f & m \\
g & f & c & n \\
l & m & n & o
\end{array}\right|=0\)

(1) Necessary Condition

Let P(x₁, y₁, z₁) be the point of contact of the given tangent plane π with the cone

E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

∴ The equation to the tangent plane is U₁x + V₁y + W₁z  = 0

⇒ (ax₁ + hy₁ + gz₁)x + (hx₁ + by₁ + fz₁ )y + (gx₁ + fy₁ + cz₁ )z = 0

Comparing with the given tangent plane π i.e. lx + my + nz = 0

We have \(\frac{\mathrm{U}_1}{l}=\frac{\mathrm{V}_1}{m}=\frac{\mathrm{W}_1}{n}\) (= -k, where k ≠ 0)

⇒ \(\frac{a x_1+h y_1+g z_1}{l}=\frac{h x_1+b y_1+f z_1}{m}=\frac{g x_1+f y_1+c z_1}{n}=-k\)

⇒ ax₁ + hy₁ + gz₁ + lk = 0; hx₁ + by₁ + fz₁ + mk = 0; gx₁ + fy₁ + cz₁ + nk = 0

Also lx₁ + my₁ + nz₁ = 0.

The non-zero solution (x₁, y₁, z₁, k) satisfy the equations

ax + hy + gz + lt = 0; hx + by + fz + mt = 0 …..(1)

gx + fy + cz + nt = 0; lx + my + nz = 0.

Hence =\(\left|\begin{array}{lllc}
a & h & g & l \\
h & b & f & m \\
g & f & c & n \\
l & m & n & o
\end{array}\right|=0 \Rightarrow p=\mathrm{A} l^2+\mathrm{B} m^2+\mathrm{C} n^2+2 \mathrm{Fmn}+2 \mathrm{G} n l+2 \mathrm{H} l m=0\)

where A, B, C, F, G, H are the cofactors of a, b, c, f, g, h in the determinant.

⇒ \(\Delta=\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\)

(2) Sufficiency of the Condition. Given ρ = 0, to prove the plane π is a tangent plane to the cone E(x, y, z) = 0.

Proof. If ρ = 0, there exists a non-zero solution (x₁, y₁, z₁, k) to I.

If k = 0, then U₁ = 0, V₁ = 0, W₁ = 0 ⇒ (x₁, y₁, z₁) is the vertex of the cone.

This contradicts the fact that (x₁, y₁, z₁) ≠ (0, 0, 0). Hence k ≠ 0.

Corresponding to the non-zero solution (x₁, y₁, z₁, k) of the equation I, we have

U₁ = -kl, V₁ = -km, W₁ = -kn …..(1) and lx₁ + my₁ + nz₁ = 0 …..(2)

(2) ⇒ P ∈ π

and E(x₁, y₁, z₁) = U₁x₁ + V₁y₁ + W₁z₁ = -k(lx₁ + my₁ + nz₁) = 0.

∴ (x₁, y₁, z₁) is a point on the cone.

∴ P(x₁, y₁, z₁) is a common point of the plane π and the cone.

∴ The equation to the tangent plane at P to the cone is U₁x + V₁y + W₁z  = 0

Since l:m:n = U₁ : V₁ : W₁, lx + my + nz = 0 is the tangent plane P(x₁, y₁, z₁) to the cone.

The Cone Reciprocal Cone

Theorem.11. The locus of the lines perpendicular to the tangent planes of the cone E(x, y, z) = ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 and passing through its vertex is the cone.

⇒ \(\left|\begin{array}{llll}
a & b & g & x \\
h & b & f & y \\
g & f & c & z \\
x & y & z & o
\end{array}\right|=0\)

Proof. Let \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\) be a line perpendicular to a tangent plane of the cone and passing through the vertex (0, 0, 0).

The cone is E(x, y, z) ≡ ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0 …..(1)

∴ lx + my + nz = 0 is the tangent plane to (1).

<=> Al² + Bm² + Cn²+ 2Fmn + 2Gnl + 2Hlm = 0

where A, B, C, F, G, H are the cofactors of a, b, c, f, g, h in the determinant.

⇒ \(\Delta=\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\)

Hence the locus of the normal line is

Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0

i.e. \(\left|\begin{array}{llll}
a & b & g & x \\
h & b & f & y \\
g & f & c & z \\
x & y & z & 0
\end{array}\right|=0\)

This equation represents a cone called the reciprocal cone of (1).

Corollary. The reciprocal cone of

Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0 …..(2)

is the cone E(x, y, z) = 0

Proof. By the above theorem, the reciprocal cone of (2) is

A’x² + B’y² + C’z²+ 2F’yz + 2G’zx + 2H’xy = 0 …..(3)

Where A’, B’, C’, F’, G’, H’ are the cofactors of A, B, C, F, G, H in determinant

⇒ \(\Delta=\left|\begin{array}{lll}
\mathrm{A} & \mathrm{H} & \mathrm{G} \\
\mathrm{H} & \mathrm{B} & \mathrm{F} \\
\mathrm{G} & \mathrm{F} & \mathrm{C}
\end{array}\right|\)

∴ A’ = BC – F² = (ca – g²)(ab – h²) – (gh – af)²

= a(abc + 2fgh – af² – bg² – ch²) = aΔ,

where Δ = abc + 2fgh – af² – bg² – ch²

Similarly B’ = bΔ, C’ = cΔ, F’ = fΔ, G’ = gΔ, H’ = hΔ

Then the equation (3) reduces to ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

i.e. E(x, y, z) = 0

Thus cones (1) and (2) are Reciprocal cones to each other.

Note 1. A cone and its reciprocal cone will have the same vertex.

2. Corresponding to each tangent plane of a cone there exists a generator of the reciprocal cone which is perpendicular to the tangent plane and vice versa.

3. A cone E(x, y, z) = 0 has three mutually perpendicular tangent planes

<=> the reciprocal cone of E(x, y, z) = 0 has three mutually perpendicular generators

<=> (bc – f²) – (ca – g²) – (ab – h²) = 0

<=> bc + ca + ab = f² + g² + h²

The Cone Solved Problems

Example.1. The semi-vertical angle of a right circular cone having three mutually perpendicular (1) generators is tan^-1√2
(2) tangent plane is \(\tan ^{-1} \frac{1}{\sqrt{2}}\)

Solution. Let the equations to the right circular cone be x² + y² = z² tan²α

(1) If the cone contains three mutually perpendicular generators then a + b + c = 0

(2) The given cone contains three mutually perpendicular tangent planes

<=> its reciprocal cone contains three mutually perpendicular generators

∴ Equations to the reciprocal cone of (1) is

-tan²αx² – tan²αy² + 1.z² = 0    …..(2)

Equation (2) will have three mutually perpendicular generators if

-tan²α – tan²α + 1 = 0 ⇒ \(\alpha=\tan ^{-1} \frac{1}{\sqrt{2}}\).

Example 2. Show that the general equation to a cone that touches the three coordinate planes is \(\sqrt{a x}+\sqrt{b y}+\sqrt{c z}=0\).

Solution. The general equation of the cone containing the three coordinates axes is ayz + bzx + cxy = 0

The reciprocal cone of (1) will have the coordinate planes.

∴ The equation to the reciprocal cone of (1) is

Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0

Where A = -a²; B = -b²; C = -c²

F = bc; G = ca; H = ab

∴ The equation to the required cone is

-a²x² – b²y² – c²z² + 2bcyz + 2cazx + 2abxy = 0

⇒ (ax + by – cz)² = 4abxy ⇒ ax + by – cz = ±2√abxy

⇒ ax + by ± 2√abxy = cz ⇒ (√ax ± √by)²= cz

⇒ √ax ± √by = ±√cz ⇒ √ax ± √by ± √cz = 0

Example.3. Show that the reciprocal cone of ax² + by² + cz² = 0 is the cone \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=0\).

Solution. Given cone is ax² + by² + cz² = 0

∴ The equation to the reciprocal cone is

Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0

Where A = bc, B = ca, C = ab, F = 0, G = 0, H = 0

∴ The equation to the reciprocal cone is

bcx² + cay² + abz² = 0 ⇒ \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=0\)

Example.4. Find the equation of the tangent planes to the cone 9x² – 4y² + 16z² = 0, which contains the line \(\frac{x}{32}=\frac{y}{72}=\frac{z}{27}\).

Solution. The given line \(\frac{x}{32}=\frac{y}{72}=\frac{z}{27}\) is the line of intersection of the planes

72x – 32y = 0 and 27y – 72x = 0

i.e., 9x – 4y = 0 and 3y – 8z = 0 …..(1)

∴ The plane passing through line (1) is 9x – 4y + λ(3y – 8z) = 0

i.e., 9x + y(3λ – 4) – 8λz = 0 …..(2)

∴ The equation to the normal line of (2) is\(\frac{x}{9}=\frac{y}{3 \lambda-4}=\frac{z}{-8 \lambda}\) …..(3)

Now plane (2) is a tangent plane to the cone 9x² – 4y² + 16z² = 0 …..(4)

<=> The normal line(3) is a generator of the reciprocal cone of the cone (4)

∴ The equation of the reciprocal cone of (4) is \(\frac{x^2}{9}-\frac{y^2}{4}+\frac{z^2}{16}=0\) …..(5)

Since (3) is a generator of (5)

⇒ \(\frac{9^2}{9}-\frac{(3 \lambda-4)^2}{4}+\frac{(-8 \lambda)^2}{16}=0 \text { i.e. } 7 \lambda^2+24 \lambda+20=0 \Rightarrow \lambda=-2 \text { or } \frac{-10}{7}\)

Hence the equations of tangent planes from (2) are 9x – 10y + 16z = 0 and 63x – 58y + 80z = 0

Example.5. Prove that the equation \(\sqrt{f x} \pm \sqrt{g y} \pm \sqrt{h z}=0\) represents a cone that touches the coordinate planes and find its reciprocal cone.

Solution. The given equation is \(\sqrt{f x} \pm \sqrt{g y} \pm \sqrt{h z}=0\)

⇒ \(f x+g y \pm 2 \sqrt{f g x y}=h z\)

⇒ \((f x+g y-h z)^2=4 f g x y\)

⇒ \(f^2 x^2+g^2 y^2+h^2 z^2-2 g h y z-2 h f x x-2 f g x y=0\) …..(1)

This being a homogenous equation of the second degree, represents a quadric cone

The coordinate plane x = 0 meets (1) in \(g^2 y^2+h^2 z^2-2 g h y z=0 \Rightarrow(g y-h z)^2=0\)

⇒ which is a perfect square

⇒ x = 0 touches it similarly we can show that y = 0, z = 0 also touch(1).

Again from the cone(1).

⇒ \(a^{\prime}=f^2, b^{\prime}=g^2, c^{\prime}=h^2, f^{\prime}=-g h,^{\prime} g^{\prime}=-h f,^{\prime} h^{\prime}=-f g\)

∴ \(\mathrm{A}=b c-f^2=g^2 h^2-(-g h)^2=0\)

Similarly \(\mathrm{B}=0, \mathrm{C}=0, \mathrm{~F}=g h-a f=(-h f)(-f g)-f^2(-g h)=2 f^2 g h\)

Similarly \(\mathrm{G}=2 g^2 h f, \mathrm{H}=2 h^2 f g\)

∴ Reciprocal cone is Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0

⇒ \(2 f^2 g h y z+2 g^2 h f z x+2 h^2 f g x y=0 \Rightarrow f y z+g z x+h x y=0\)

Example.6. Find the condition that one plane ux + vy + wz = 0 may touch the cone ax² + by² + cz² = 0

Solution. Equation to the normal to the given plane is \(\frac{x}{u}=\frac{y}{v}=\frac{z}{w}\) …..(1)

The equation to the reciprocal cone of

⇒ \(a x^2+b y^2+c z^2=0 \text { is } \frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=0\) …..(2)

Now the plane touches the cone (2)

<=> The normal of the plane lies on cone (2)

<=> \(\frac{u^2}{a}+\frac{v^2}{b}+\frac{w^2}{c}=0\) which is the required condition.

Example.7. Find the equation of the cone that touches the three coordinate planes and the planes x + 2y + 3z = 0, 2x + 3y + 4z = 0.

Solution. The equation to the cone touching the three axes can be taken as \(\sqrt{f x} \pm \sqrt{g y} \pm \sqrt{h z}=0\)

Its reciprocal cone is fyz + gzx + hxy = 0 …..(2)

The planes x + 2y + 3z = 0 and 2x + 3y + 4z = 0

touch the cone (1) <=> their normals lies on (2)

⇒ D.r’s of the normal (1, 2, 3) and (2, 3, 4) satisfy (2)

(1) f(2)(3) + g(3)(1) + h(1)(2) = 0 ⇒ 6f + 3g + 2h = 0 …..(3)

(2) f(3)(4) + g(4)(2) + h(2)(3) = 0 ⇒ 6f + 4g + 3h = 0 …..(4)

Solving (3) and (4): \(\frac{f}{9-8}=\frac{g}{12-18}=\frac{h}{24-18}\)

Hence (1) becomes \(\sqrt{x}+\sqrt{-6 y}+\sqrt{6 z}=0\)

The Cone Intersection Of Two Cones With A Common Vertex

In general two cones with a common vertex intersect along four common generators.

Let S = 0 and S’ = 0 be two cones with origin as the common vertex, then S + λS’ = 0 represents the general equation of a cone whose vertex is at the origin and which passes through the four common generators of the two cones.

Cor. Let λ be so chosen that S + λS’ = 0 becomes the product of two linear factors the two linear factors equated to zero represent the equations to a pair of planes through the common generators.

In that case the values of λ are the roots of the λ – cubic eqution

⇒ \(\left|\begin{array}{lll}
a+\lambda a^{\prime} & h+\lambda h^{\prime} & g+\lambda g^{\prime} \\
h+\lambda h^{\prime} & b+b^{\prime} \lambda & f+\lambda f^{\prime} \\
g+\lambda g^{\prime} & f+\lambda f^{\prime} & c+\lambda c^{\prime}
\end{array}\right|=0\)

The three values of λ give the three pairs of planes through the four common generators.

The Cone Solved Problems

Example.1. Find the equation of the cone which passes through the common generators of the cones 2x² – 4y² – z² = 0 and 10xy – 2yz + 5zx = 0 may the line with direction ratios (1, 2, 3).

Solution.

Given

2x² – 4y² – z² = 0 and 10xy – 2yz + 5zx = 0

Let the required cone be 2x² – 4y² – z² + λ(10xy – 2yz + 5zx) = 0 …..(1)

This is a quadric cone with a vertex at the origin.

The line with d.r.’s (1, 2, 3) lies on (1)

<=> 2(1)² – 4(2)² – (3)² + λ[10(1)(2) – 2(2)(3) + 5(3)(1)] = 0

⇒ – 23 + 23λ = 0 ⇒ λ = 1

∴ Required cone is 2x² – 4y² – z² + 10xy – 2yz + 5zx = 0

Example 2. Find the condition that the lines of the section of the plane lx + my + nz = 0 and the cones ax² + by² + cz² = 0 and fyz + gzx + hxy = 0 should be concident.

Solution. Any cone through the intersection of the two given cones is ax² + by² + cz² + λ(fyz + gzx + hxy) = 0 …..(1)

Given that the plane lx + my + nz = 0 cuts (1) in coincident lines

⇒ for some value of λ(1) must represent a pair of planes.

Let l₁x + m₁y + n₁z = 0 …..(2) be the other plane.

Then ax² + by² + cz² + λ(fyz + gzx + hxy) = (lx + my + nz)(l1x + m1y + n1z)

⇒ ll₁ = a, mm₁ = b, nn₁ = c

⇒  l₁ = a/1, m₁ = b/m, n₁ = c/n

Again \(\lambda f=m n_1+m_1 n=\frac{c m}{n}+\frac{b n}{m}=\frac{c m^2+b n^2}{m n}\)

Similarly \(\lambda g=\frac{a n^2+c l^2}{n l} \text { and } \lambda h=\frac{a m^2+b l^2}{l m}\)

⇒ \(\frac{c m^2+b n^2}{f m n}=\frac{a n^2+c l^2}{g n l}=\frac{a m^2+b l^2}{h l m}\) which is the required condition.

Chapter 8 Cyclic Groups 8.1 Before Defining A Cyclic Group, We Prove A Theorem That Serves As A Motivation For The Definition Of Cyclic Group.

Theorem. Let G be a group and a be an element of G. Then \(\left.\mathbf{H}=\left\{a^n\right\rangle \in \mathbf{Z}\right\}\) is a subgroup of G. Further H is the smallest of subgroups of G which contain the element a.

Proof. Let (G, .) be a group and a ∈ G .

For 1 ∈ Z we have \(a^{\prime}=a \in \mathbf{H}\) which shows that H is nonempty.

Suppose now that \(a^r, a^s \in \mathbf{H}\). We will show that

1)\(a^r a^s \in \mathbf{H}\) and \(\left(a^r\right)^{-1}\) \in \mathbf{H}[/latex] which will prove that H is a subgroup of G.

⇒ \(a^r, a^s \in \mathbf{H} \Rightarrow r, s \in \mathbf{Z} \Rightarrow r+s,-r \in \mathbf{Z}\)

∴ \(a^r \cdot a^s=a^{r+s} \in \mathbf{H} \text { and }\left(a^r\right)^{-1}=a^{-r} \in \mathbf{H}\)

∴ H is a subgroup of G.

2)Suppose K is any other subgroup of G such that a ∈ K

Then \(a^n \in \mathbf{K} \forall n \in \mathbf{Z}\).

∴ \(\mathbf{H} \subset \mathbf{K}\) which shows that H is the subset of every subgroup of G which contains a.

Thus H is the smallest of subgroups of G which contain a.

Chapter 8 Cyclic Groups 8.2 Cyclic Subgroup Generated By a

Definition. Suppose G is a group and a is an element of G. Then the subgroup \(\mathbf{H}=\left\{a^n / n \in \mathbf{Z}\right\}\) is called a cyclic subgroup generated by a. a is called a generator of H.

This will be written as H = < a > or (a) or {a}.

Cyclic group.

Definition. Suppose G is a group and there is an element a ∈ G such that \(\mathbf{G}=\left\{a^n \mid n \in \mathbf{Z}\right\}\) Then G is called a cyclic group and a is called a generator of G.

We denote G by < a >.

Thus a group consisting of elements that are only the power of a single element belonging to it is a cyclic group.

Let G be a group and a ∈ G: If the cyclic subgroup of G generated by a i.e. < a > is finite, then the order of the subgroup i.e. \(\mid<a>\mid\) is the order of a. If < a > is infinite then we say that the order of a is infinite.

Note. If G is a cyclic group generated by a, then the elements of G will be \( \ldots a^{-2}, a^{-1}, a^0=e, a^1, a^2, \ldots\) in multiplicative notation and the elements of G will be\(\ldots-2 a,-a, 0 a=0, a, 2 a, \ldots\)in additive notation. The elements of G are not necessarily distinct. There exist finite and infinite cyclic groups.

Definition Of Cyclic Groups With Examples

Example 1. G = {1,-1} is a multiplicative group. Since \((-1)^0=1,(-1)^1=-1,(\mathbf{G}, .)\) is a cyclic group generated by – 1 i.e. G = < -1 > • It is a finite cyclic group of order 2 and 0(-l) = 2.

Example 2. G = {…- 4, – 2,0,2,4,….} is an additive group.

Since G = \(\mathbf{G}=\{2 m / m=\ldots-1,0,1,2, \ldots\}\), G is a cyclic group generated by 2 i.e.

G = < 2 > . It is an infinite cyclic group.

Example 3. (Q, +), \((Q^+, •)\) are groups but are not cyclic.

Example 4. \(\left\{12^n / n \in \mathbf{Z}\right\}\) is a cyclic group w.r.t. usual multiplication.

Its generators are 12,1/12.

Example 5. <18> is a cyclic subgroup of the cyclic group \(\left(\mathbf{Z}_{36},+_{36}\right)\), and since

⇒ \(18^1=18,18^2=18+{ }_{36} 18=0\)

⇒ \(18^3=18^2+_{36} 18=18,18^4=18^3++_{36} 18=18+{ }_{36} 18=0, \ldots \ldots,\) ,we have < 18 > = {0,18}.

Example 6. Let \(\mathbf{G}=\mathbf{S}_3\) and H = {(1),(13)}.

Then the left cosets of H in G are :

(1) H = H,(l 2) \(H = \left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 3
\end{array}\right)\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right)=\left(\begin{array}{lll}
1 & 3 & 2
\end{array}\right) \mathbf{H}\)

(1 3)H = H,(2 3)H = (1 2 3)H. (Analogous results hold for right cosets)

Theorem 1. Let (G, .) be a cyclic group generated by a. If O(a) = n, then \(a^n=e \text { and }\left\{a, a^2, \ldots . a^{n-1}, a^n=e\right\}\) is precisely the set of distinct elements belonging to G where e is the identity in the group (G)

Theorem 2. If G is a finite group and \(a \in \mathbf{G} \text {, then } o(a) / o(\mathbf{G})\)

Proof. G is a finite group. Let o (G) = m .

Let H be the cyclic subgroup of G generated by a.

Let o (a) = n

∴ o (H) = n (Vide Theorem 25, Art. 2.17.)

But by Lagrange’s Theorem, o (H) / o (G).

n / o(G) i.e. o (a) / o(G).

Note. If o(a) = n and a ∈ G , then \(o(\mathbf{H}) \leq o(\mathbf{G})\).

Theorems On Cyclic Groups With Proofs And Examples

Theorem 3. If G is a finite group of order n and if a ∈ G, then \(a^n=e\) (identity in G).

Given

If G is a finite group of order n and if a ∈

Proof. Let o (a) = d and \(a^d=e \text { and } d \leq n\).

If H is a cyclic subgroup generated by a, then o (H) = d = o (a).

But by Lagrange’s Theorem, o (H) / o (G) i.e. d/n.

∴ ∃ a positive integer q such that n = dq.

∴ \(a^n=a^{d q}=\left(a^d\right)^q=e^q=e\)

Note. the statement of the above theorem may be: If G is a finite group, then for any \(a \in \mathbf{G}, a^{0(G)}=e\).

Cyclic Groups Solved Problems

Example 1. Let \(\mathbf{A}=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right], \mathbf{C}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right] \text { and } \mathbf{D}=\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\).
We have that G = {A, B, C, D} with matrix multiplication as operation is a group whose composition table is given below. Show that G is a cyclic group with generator B.

Solution: here O(G) = 4, A is the identity element in G. Now we can see that

⇒ \(\mathbf{B}^1=\mathbf{B}, \mathbf{B}^2=\mathbf{B} \cdot \mathbf{B}=\mathbf{C}\)

⇒ \(\mathbf{B}^3=\mathbf{B}^2 \cdot \mathbf{B}=\mathbf{C} \cdot \mathbf{B}=\mathbf{D}\)

⇒ \(\mathbf{B}^4=\mathbf{B}^3 \cdot \mathbf{B}=\mathbf{D} \cdot \mathbf{B}=\mathbf{A}\)

Thus B ∈ G generated the group G and hence G is a cyclic group with B as a generator

i.e. G=<B>.

Note that O(B)=4=O(G) and \(\mathbf{B}^{\mathbf{O}(\mathbf{G})}=\mathbf{A}\).

Also, G is abelian.

Note: A, C, and D are not generators of group G.

Example 2. Prove that (Z, +) is a cyclic group.

Solution: (Z, +) is a group and 1 ∈ Z.

When we take additive notation in Z, a becomes na.

⇒ \(1^0=0.1=0,1^1=1.1=1,1^2=2.1=2\) etc.

Also \(1^{-1}=-1,1^{-2}=-2.1=-2\) etc.

∴ 1 is a generator of the cyclic group (Z, +) i.e. Z = <1>.

Similarly, we can prove that Z = < -1 > .

Note 1. (Z, +) has no generators except 1 and – 1.

For: Let r = 4 ∈ Z.

We cannot write every element m of z in the form m = 4n. For example, 7 = 4n is not possible for n ∈ Z. Thus when r is an integer greater than 1, r is not a generator of Z.

Similarly, when r is an integer less than – 1 also, r is not a generator of Z.

Thus (Z, +) is a cyclic group with only two generators 1 and – l.

2. (Z,- +) is an infinite abelian group and it is a cyclic group.

Classification Of Cyclic Groups With Detailed Explanation

Example 3. Show that G = {1, -1, i, -i} the set of all fourth roots of unity is a cyclic group w.r.t. multiplication.

Solution.

Given

G = {1, -1, i, -i}

Clearly (G,.) is a group. We see that

⇒ \((i)^1=i, i^2=i, i=-1, i^3=i^2 \cdot i=-1 . i=-i\)

⇒ \(i^4=i^3 \cdot i=(-i) i=1\)

Thus all the elements of G are the powers of i ∈ G i.e. G =< i >

Similarly, we can have G =< -i >. Note that 0 (G) = 0 (i) = 0 (-i) = 4 .

Also, G is abelian.

Note. (G, .) is a finite abelian group which is cyclic.

Example 4. Show that the set of all cube roots of unity is a cyclic group w.r.t. multiplication.

Solution. If ω is one of the complex cube roots of unity, we know that G = \({l,ω, ω^2}\) is a group w.r.t. multiplication. We see that \(\omega^1=\omega, \omega^2=\omega \omega=\omega^2, \omega^3=1\) .

∴ Then elements of G are the powers of the single element ω ∈ G

∴ G = <ω>

We can also have \(\mathbf{G}=\left\langle\omega^2\right\rangle\).

(∵ \(\left.\left(\omega^2\right)^1=\omega^2,\left(\omega^2\right)^2=\omega,\left(\omega^2\right)^3=1\right)\)

Example 5. Prove that the group \( \left(\{1,2,3,4\}, \times_5\right)\) is cyclic and. write its generators.

Solution. \(2 x_5 2=4,2 \times_5 2 x_5 2=4 x_5 2=3,2 \times_5 2 x_5 2 x_5 2=1,2 x_5 2 x_5 2 x_5 2 x_5 2=2\)

⇒ 2 is a generator of the group ⇒ the group is cyclic.

Also group 3 is a generator.

1, 4 are not generators of the cyclic group.

Example 6. Show that \(\left(\overline{\mathbf{Z}}_5,+\right)\)where \(\overline{\mathbf{Z}}_S\) is the set of all residue classes modulo 5, is a cyclic group w.r.t. addition (+) of residue classes.

Solution. The composition table for the group \(\left(\overline{\mathbf{Z}}_5,+\right)\) is;

We can have

⇒ \(\overline{1}=\overline{1},(\overline{1})^2=2(\overline{1})=\overline{1}+\overline{1}=\overline{2}\)

⇒ \((\overline{1})^3=\left(\overline{1}^2\right)+(\overline{1})=\overline{2}+\overline{1}=\overline{3}\)

⇒ \((\overline{1})^4=(\overline{1})^3+(\overline{1})=\overline{3}+\overline{1}=\overline{4}\)

Thus \(\left(\overline{\mathbf{Z}}_5,+\right)\) is a cyclic group with 1 as generator.

∴ \(\left(\overline{\mathbf{Z}}_5,+\right)=\langle\overline{1}\rangle\)

Similarly, we can prove that 2,3,4 are also generators of this cyclic group.

Example 7. Show that \(\left(\overline{\mathbf{Z}}_m,+\right)\) , where \(\overline{\mathbf{z}}_m\)is the set of all residue classes modulo m and + is the residue class addition, is a cyclic group.

Solution. We have \(\left(\overline{\mathbf{Z}}_m,+\right)\) as an abeian group.

We can have \((\overline{1})^1=1,(\overline{1})^2=2(\overline{1})=\overline{1}+\overline{1}=\overline{2}\)

… … … … …

⇒ \((\overline{1})^{m-1}=(\overline{1})^{m-2}+(\overline{1})^1=(m-2) \overline{1}+\overline{1}=\overline{m-1}\)

We have nZ = {0, ± n, ± 2n, ± 3n, …}

∴ (nZ, +) = <n > and (nZ, +) = <-n >.

Note. (nZ, +) is a subgroup of the group (Z, +) and (nZ, +) is cyclic.

Cyclic Groups 8.3. Some Properties Of Cyclic Groups

Theorem 4. Every cyclic group is an abelian group.

Proof. Let G = < a > by a cyclic group.

We have \(\mathbf{G}=\left\{a^n / n \in \mathbf{Z}\right\}\)

Let \(a^r, a^s \in \mathbf{G}\)

∴ \(a^r \cdot a^s=a^{r+s}\)

= \(a^{s+r}=a^s \cdot a^r\)

G is abelian.

Note. Converse is not true i.e. every abelian group is not cyclic. Klein’s group of 4 is an example.

e = \(e^2=e^3=e^4 ; a=a, a^2=e, a^3=a, a^4=e\)

b = \(b, b^2=e, b^3=b, b^4=e, c=c, c^2=e, c^3=c, c^4=e\)

None of the elements of G generates G even though G is abelian

i.e. G is abelian but not cyclic.

example. Consider the set G = {A, B, C, D} where

⇒ \(\mathbf{A}=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}
-1 & 0 \\
0 & 1
\end{array}\right], \mathbf{C}=\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right], \mathbf{D}=\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right]\)

and the matrix multiplication as the binary composition on G.

Composition table is :

Clearly, G is a finite abelian group (of order 4) with identity element A.

Also \(\mathbf{B}^2=\mathbf{A}, \mathbf{C}^2=\mathbf{A} \text { and } \mathbf{D}^2=\mathbf{A}\)

i.e. each element is of order 2 (except the identity A )

∴ G is abelian.

Hence there is no element of order 4 in G.

∴ G is not cyclic and hence every finite abelian group is not cyclic.

Solved Problems On Cyclic Groups Step-By-Step

Theorem 5. If a is a generator of a cyclic group G, then \(a^{-1}\) is also a generator of G.

If \(\mathbf{G}=\langle a\rangle \text { then } \mathbf{G}=\left\langle a^{-1}\right\rangle\).

Proof. Let G = <a> be a cyclic group generated by a. Let \(a^r \in \mathbf{G}, r \in \mathbf{Z}\)

We have \(a^r=\left(a^{-1}\right)^{-r} \text { since }-r \in \mathbf{Z}\)

∴ Each element of G is generated by \(a^{-1}\)

Thus \(a^{-1}\) is also a generated of G

i.e., \(\mathrm{G}=\left\langle a^{-1}\right\rangle\)

Theorem 6. Every subgroup of the cyclic group is cyclic.

Proof. Let G =<a>. Let H be a subgroup of G. Since H is a subgroup of G, we take that every element of H is an element of G. Thus it can be expressed as \(a^n\) for some n ∈ Z.

Let d be the smallest of the positive integers such that \(a^n \in \mathbf{H}\). We will now prove that \(\mathbf{H}=\left\langle a^d\right\rangle\).

Let \(a^m \in \mathbf{H} \text { where } m \in \mathbf{Z}\)

By division algorithm, we can find integers g and r such that

m = dq + r 0 < r < d .

∴ \(a^m=a^{d q+r}=a^{d q} a^r=\left(a^d\right)^q \cdot a^r\)

But \(a^d \in \mathbf{H} \Rightarrow\left(a^d\right)^q \in \mathbf{H} \Rightarrow a^{d q} \Rightarrow a^{-d q} \in \mathbf{H}\)

Now \(a^m, a^{-d q} \in \mathbf{H} \Rightarrow a^m \cdot a^{-d q} \in \mathbf{H} \Rightarrow a^{\mu n-d q} \in \mathbf{H} \quad \Rightarrow a^r \in \mathbf{H}\)

But 0 < r < d and \(a^r \in \mathbf{H}\) is a contradiction to our assumption that of smallest integer such that \(a^d \in \mathbf{H}\) .

∴ r = 0

∴ m = dq

i.e. \(a^m=\left(a^d\right)^q\) which shows that every \(a^n \in \mathbf{H}\) can be written as \(\left(a^d\right)^q, q \in \mathbf{Z}\).

∴ \(\mathbf{H}=\left\langle a^d\right\rangle\)

Hence a subgroup LI of G is cyclic and ad is a generator of H

Note: The converse of the above theorem is not true.

That is though the subgroup of a group is cyclic, the group need not be cyclic.

We know that (Z, +) is a subgroup of (R, +) . We also have that (Z, +) is a cyclic group generated by 1 and – 1. But (R, +) is not cyclic. group since it has no generators.

Cor. Every subgroup of a cyclic group is a normal subgroup.

Proof. Every cyclic group is abelian (vide Theorem 4) and every subgroup of a cyclic group is cyclic (vide Theorem 6).

∴ Every subgroup of a cyclic group is abelian. Hence every subgroup of a cyclic group is a normal subgroup.

Cyclic Groups 8.4. Classification Of Cyclic Groups

Let G =<a>. Then

1)G is a finite cyclic group if there exist two unequal integers l and m such that \(a^{\prime}=a^m\)

If a group G of order n is cyclic, then G is a cyclic group of order n.

2)G is an infinite cyclic group if for every pair of unequal integers l and m, \(a^l \neq a^m\)

Theorem 7. The quotient group of a cyclic group is cyclic.

Proof. Let G =<a> be a cyclic group with a as a generator.

Let N be a subgroup of G. Since G is abelian we take that N is normal in G.

We know that \(\frac{\mathbf{G}}{\mathbf{N}}=\{\mathbf{N} x / x \in \mathbf{G}\}\)

Now \(a \in \mathbf{G} \Rightarrow \mathbf{N} a \in \mathbf{G} / \mathbf{N} \Rightarrow\langle\mathbf{N} a\rangle \subseteq \mathbf{G} / \mathbf{N}\)

Also \(\mathbf{N} x \in \mathbf{G} / \mathbf{N} \Rightarrow x \in \mathbf{G}=\langle a\rangle\)

∴ \(x=a^n\) for some n ∈ Z.

∴ Nx = Na = N(a a a ……n times) when n is a +ve integer

= NaNa……n times = \((\mathbf{N} a)^n\)

We can prove that \(\mathbf{N} x=(\mathbf{N} a)^n\) when n = 0 or a negative integer.

∴ \(\mathbf{N} x \in \mathbf{G} / \mathbf{N} \Rightarrow \mathbf{N}(x) \in\langle\mathbf{N} a\rangle\)

∴ \(\mathbf{G} / \mathbf{N} \subseteq\langle\mathbf{N} a\rangle\)

∴ From (1) and (2) G/N = <Na> which shows that G/N is cyclic.

Theorem 8. If p is a prime number then every group of order p is a cyclic group i.e. a group of prime order is cyclic.

Proof. Let p ≥ 2 be a prime number and G be a group such that O(G) ≥ p.

Since the number of elements is at least 2, one of the elements of G will be different from the identity e of G. Let that element be a.

Let <a > be the cyclic subgroup of G generated by a.

∴ \(a \in<a>\Rightarrow<a>\neq\{e\}\)

Let <a> have order h,

∴ By Lagrange’s Theorem h \ p

But p is a prime number.

∴ h = 1 of h = p

But \(<a>\neq\{e\}\)

∴ h ≠ 1 and hence h = p

∴ O( <a>) = p i.e. <a>= G which shows that G is a cyclic group.

Note 1. We have by the above theorem if O(G) – p, a prime number, then every element of G which is not an identity is a generator of G. Thus the number of generators of G having p elements is equal to p – 1.

2. Every group G of orders less than 6 is abelian. For: We know that every group G of order less than or equal to 4 is abelian.

Also, we know that every group of prime order is cyclic and every cyclic group is abelian. If O(G) = 5, then G is abelian.

Thus the smallest non-abelian group is of order 6.

3. Is the converse of the theorem “Every group of prime order is cyclic” true? Not true.

For 4th roots of unity w.r.t. multiplication forms a cyclic group and 4 is not a prime number. Thus a cyclic group need not be of prime order.

Cyclic Groups 8.5 Some More Theorems on Cyclic Groups

Theorem 9. If a finite group of order n contains an element of order n, then the group is cyclic.
Proof. Let G be a finite group of order n. Let a ∈ G such that O (a) = n

i.e. \(a^n=e\) where n is the least positive integer.

If H is a cyclic subgroup of G generated by a i.e. if \(\mathbf{H}=\left\{a^r / r \in \mathbf{Z}\right\}\) then O (H) = n because the order of the generator a of H is n.

Thus H is a cyclic subgroup of G and O (H) = 0 (G).

Hence H = G and G itself is a cyclic group with a as a generator.

Note. Suppose G is a finite group of order n and we are to determine whether G is cyclic or not.

For this we find the orders of the elements of G and if a ∈ G exists such that O (a) = n then G will be a cyclic group with a as a generator.

Exercises On Cyclic Groups With Solutions

Theorem 10. Every finite group of composite order possesses proper subgroups.
Proof. Let G be a finite group of composite order mn where m(≠) and n(≠) are positive integers.

1)Let G = <a>. Then O (a) = O (G) = mn

∴ \(a^{m n}=e \Rightarrow\left(a^n\right)^m=e \Rightarrow \mathrm{O}\left(a^n\right) \text { is finite and } \leq m\)

Let \(\mathrm{O}\left(a^n\right)=p\). where p < m.

Then \(\left(a^n\right)^p=e \Rightarrow a^{n p}=e\)

But p < m ⇒ np < mn

Thus \(a^{n p}=e\) where np < mn .

Since O (a) = mn,\(a^{n p}=e\) is not possible, so p = m

∴ \(\mathbf{O}\left(a^n\right)=m\)

∴ \(\mathbf{H}=\left\langle a^n\right\rangle\) is a cyclic subgroup of G and \(\mathbf{O}(\mathbf{H})=\mathbf{O}\left(a^n\right)\)

Thus O (H) = m .

Since 2 ≤ m ≤ n, H is a proper cyclic subgroup of G.

2) Let G be not a cyclic group.

Then the order of each element of G must be less than mn. So there exists an element, say b in G such that 2 ≤ O (b) < mn. Then H = <b > is a proper subgroup of G

Theorem 10(a). If G is a group of order p,q are prime numbers, then every proper subgroup of G is cyclic.

Proof. Let H be a proper subgroup of G where | G | = pq (p,q are prime numbers)

By Lagrange’s Theorem, | H | divides | G |.

∴ Either | H | = 1 or p or q.

∴| H | = 1 ⇒ H = (e) which is cyclic;

| H | = P (P is Prime) ⇒ H is cyclic and | H | = q {q is prime) ⇒ H is cyclic.

∴ H is a proper subgroup of G which is cyclic. Hence every proper subgroup of G is cyclic.

Theorem 11. If a cyclic group G is generated by an element a of order n, then \( a^m\) is a generator of G iff the greatest common divisor of m and n is 1 i.e. iff m,n are relatively prime i.e. (m,n)=1.

Proof. Let G = <a> such that O (a) = n i.e. \(a^n=e\).

Group G contains exactly n elements.

1)Let m be relatively prime to n. Consider the cyclic subgroup \(\mathbf{H}=\left\langle a^m\right\rangle\) of G.

Clearly H⊆G…(l) . since each integral power of am will be some integral power of a.

Since m and n are relatively prime, there exist two integers x and y such that mx + ny = 1.

∴ \(a=a^1=a^{m x+n y}=a^{n x} \cdot a^{n y}=a^{m x} \cdot\left(a^n\right)^y \cdot=a^{m x} e^y=a^{m x} e=\left(a^m\right)^x\)

∴ Each integral exponent of a will also be some integral exponent of \(a^m\).

∴ G ⊆ H

∴ From (1) and (2), H = G and \(a^m\) is a generator of G.

2)Let \(\mathbf{G}=\left\langle a^m\right\rangle\). Let the greatest common divisor of m and n be d(≠1) i.e. d > 1. Then m/d, n/d just be integers.

Now \(\left(a^m\right)^{n / d}=a^{m n / d}=\left(a^n\right)^{m / d}=e^{m / d}=e\)

∴ \(\mathbf{O}\left(a^m\right)<n\)

(∵ \(\frac{n}{d}<n\))

∴ \(a^m\)cannot be a generator of G because the order of am is not equal to the order of G. So d must be equal to 1. Thus m and n are relatively prime.

Note 1. If G = < a > is a cyclic group of order n, then the total number of generators of G will be equal to the number of integers less than and prime to n.

2. \(\mathbf{Z}_8\) is a cyclic group with 1,3,5,7 as generators.

Note that < 3 > = (3, (3 + 3) mod 8, (3 + 3 + 3) mod 8,……} = (3,6,1,4,7,2,5,0} = \(\mathbf{Z}_8\)

< 2 > = {0,2,4,6} ≠ \(\mathbf{Z}_8\) implies 3 is a generator and 2 is not a generator of \(\mathbf{Z}_8\).

Theorem 12. If G is a finite cyclic group of order n generated by a, then the subgroups of G are precisely the subgroups generated by \( a^m\) where m divides n.

Proof. Since G is a finite cyclic group of order n generated by a, then \(a^m\) generates a cyclic subgroup, say H of G.

Since O(G) = .n, \(a^n=e\) where e is the identity in G.

Since H is a subgroup of G,e ∈ H i.e.\(a^n \in \mathbf{H}\).

If m is the least positive integer such that \(a^m \in \mathbf{H}\) then by division algorithm there exist positive integers q and r such that \(n=m q+r, \quad 0 \leq r<m\).

∴ \(a^n=a^{m q+r}=a^{m q} \cdot a^r=\left(a^m\right)^q \cdot a^r\)

But \(a^m \in \mathbf{H}\) .

∴ \(\left(a^m\right)^q \in \mathbf{H} \Rightarrow a^{m q} \in \mathbf{H} \Rightarrow a^{-m q} \in \mathbf{H}\).

Now \(a^n \in \mathbf{H} \Rightarrow a^{-m q} \in \mathbf{H} \Rightarrow a^{n-m q} \in \mathbf{H} \Rightarrow a^r \in \mathbf{H}\)•

But o < r < m and \(a^r \in \mathbf{H}\) is a contradiction to our assumption that m is the smallest positive integer such that \(a^m \in \mathbf{H}\)

∴ r = 0.

∴ n = mq i.e. m divides n and \(a^n=a^{m q}=\left(a^m\right)^q \in \mathbf{H}\) which means that \(a^m\) generates the cyclic subgroup H of G.

Example 10. Find all orders of subgroups of \(Z_6, Z_8, Z_{12}, Z_{60}\)

Solution. \((Z_6,+6)\) is a cyclic group and its subgroups have orders 1, 2, 3, 6 (Theorem.12)

(Proper subgroup of order 2 is ({0,3},+6),

A proper subgroup of order 3 is ({0,2,4},+6)).

⇒ \(\left(\mathbf{Z}_8,+_8\right)\) is a cyclic group and its subgroups have orders 1, 2, 4, 8.

⇒ \(\left(\mathbf{Z}_{12},+_{12}\right)\) is a cyclic group and its subgroups have orders 1,2, 3, 4, 6,12.

⇒ \(\left(\mathbf{Z}_{60},+_{60}\right)\) is a cyclic group and its subgroups have orders 1,2,3,4,5,6,10,12,15,20,30,60.

Example 11. Write clown all the subgroups of a finite cyclic group G of order 18, the cyclic group being generated by a.

Solution. Let e be the identity in G = < a >.

Now G, {e} are the trivial subgroups of G and generated by a and a18 =e respectively.

The other proper subgroups are precisely the subgroups generated by am where m divides 18. Such m’s are 2, 3, 6, 9. These subgroups are

⇒ \(\left.\left\langle a^2\right\rangle=\left\{a^2, a^4, a^6, a^8, a^{10}, a^{12}, a^{14}, a^{16}, a^{18}=e\right\},<a^3\right\rangle=\left\{a^3, a^6, a^9, a^{12}, a^{15}, a^{18}=e\right\}\)

⇒ \(\left\langle a^6\right\rangle=\left\{a^6, a^{12}, a^{18}=e\right\},\left\langle a^9\right\rangle=\left\{a^9, a^{18}=e\right\}\)

Understanding Cyclic Groups In Abstract Algebra

Theorem 13. The order of a cyclic group is equal to the order of its generator.

Proof. Let G be a cyclic group generated by a i.e. G =<a>

1) Let O(n)= n, a a finite integer.

Then \(e=a^0, a^1, a^2, \ldots \ldots, a^{n-1} \in \mathbf{G}\)

Now we prove that these elements are distinct and these are the only elements of G such that O(G) = n.

Let i,j ( ≤ n – 1) be two non-negative integers such that \(a^i=a^j \text { for } i \neq j\)

Now either i>j or i<j.

Suppose i > j. Then \(a^{i-j}=a^{j-j} \Rightarrow a^{i-j}=a^0=e \text { and } 0<i-j<n\)

But this contradicts the fact that O (a) = n. Hence i=j

∴ \(a^0, a^1, a^2, \ldots \ldots, a^{n-1}\) are all distinct.

Consider any \(a^p \in \mathbf{G}\) where p is any integer.

By Euclid’s Algorithm, we can write p = nq. + r for some integers q and y such that 0 ≤ r < n.

Then \(a^p=a^{n q+r}=\left(a^n\right)^q \cdot a^r=e^q \cdot a^r=e \cdot a^r=a^r\)

But \(a^r\) is one of \(a^0, a^1, \ldots \ldots, a^{n-1}\)

Hence each \(a^p \in \mathbf{G}\)is equal to one of the elements \(a^0, a^1, \ldots \ldots, a^{n-1}\)

i.e. O(G) = n = o{a).

2) LetO(a) be infinite. .

Let m, n be two positive integers such that \(a^m=a^n \text { for } n \neq m\).

Suppose m > n . Then \(a^{m-n}=a^0=e\) ⇒ O(a) is finite.

It is a contradiction to the fact that O (a) is infinite.

∴ n = m i.e. for every pair of unequal integers m and n , \(a^m \neq {a}^n\)

Hence G is of infinite order.

Thus from (1)and(2), the order of a cyclic group is equal to the order of its generator. .

Note. Thus : Let (G,.) be a group and a ∈ G

If a has finite order, say, n, then <a> = \(\left\{e, a, a^2, \ldots \ldots ., a^{n-1}\right\}\) and \(a^i=a^j\) if and only if n divides i-j .

If a has infinite order, then all distinct powers of a are distinct group elements.

Theorem. 14. If G is a cyclic group of order n, then there is a one-to-one correspondence between the subgroups of G and positive divisors of n.

Proof. Let G = <a> be a finite cyclic group of order n.

∴ O(a) = n, a +ve integer. If d (a +ve integer) is a divisor of n, ∃ a +ve integer m such that n = dm.

Now O (a) = n ⇒ \(a^n=e \Rightarrow a^{d m}=e \Rightarrow\left(a^m\right)^d=e \Rightarrow \mathbf{O}\left(a^m\right) \leq d\)

Let \(O (a^m) = s\) where s < d.

Then \((a^m)s\) = e ⇒ \(a^ms = e\) where ms < md i.e. ms < n .

Since 0(a) = n, when ms < n , \(a^ms = e\) is absurd.

∴ i.e. s < d s=d.

∴ \(a^m \in \mathbf{G} \text { where } \mathbf{O}\left(a^m\right)=d\)

Thus \(<a^m>\) is a cyclic subgroup of order d.

Now we show that \(<a^m >\) is a unique cyclic subgroup of G of order d.

We know that every subgroup of a cyclic group is cyclic. If possible suppose that there is another subgroup \(<a^k>\) of G of order d where n = dm.

We shall have to show that \(<a^k> = <a^m>\).

By division algorithm ∃ integers q and r such that k = mq + r where 0 ≤ r < m …(1)

∴ kd = mqd + rd where 0 ≤ rd < md

Now \(a^{k d}=a^{m q d+r d}=a^{m q d} \cdot a^{r d}=\left(a^{m d}\right)^q \cdot a^{r d}=\left(a^n\right)^q \cdot a^{r d}=e^q \cdot a^{r d}=e \cdot a^{r d}\)

⇒ \(a^{k d}=a^{r d}\) …(2)

Since \(<a^k>\) is of order d, \(0(a^k)\) = d

e = \(a^kd\) = e.

⇒ \(a^rd = e\) from (2) which is impossible (∵ rd < md ⇒ rd < m) unless r = 0 .

∴ From (1), k = mq ⇒ \(a^k = a^mq – (a^m)^q\) ⇒ \(a^k \in\left\langle a^m\right\rangle \Rightarrow\langle a\rangle \subseteq\left\langle a^m\right\rangle\)

But number of elements in \(<a^k >\) = number of elements in \(<a^m>\) .

∴ \(\left\langle a^k\right\rangle=\left\langle a^m\right\rangle\)

∴If G is a finite cyclic group of order n, there corresponds a unique subgroup of G of order d for every divisor d of n i.e. there is a 1 – 1 correspondence between the subgroups of G and positive divisors of n.

[∵ a one-on-one mapping is always possible to be defined between the set of subgroups of order d (any +ve divisor of n) and the set of +ve divisors of n]

Classification Of Finite And Infinite Cyclic Groups With Examples

Theorem 15. Every isomorphic image of a cyclic group is again cyclic.

Proof. Let G be a cyclic group generated by a so that \(a^n\) ∈ G from n ∈ Z

Let G’ be its isomorphic image under an isomorphism f.

Now \(a^n \in \mathbf{G} \Rightarrow f\left(a^n\right) \in \mathbf{G}^{\prime}\)

∴ \(f\left(a^n\right)=f(a \cdot a \cdot a, \ldots \ldots, n \text { times })\) when n is a +ve integer

= \(f(a) \cdot f(a) \ldots \ldots, n \text { times }=[f(a)]^n\)

We can prove that \(f\left(a^n\right)=[f(a)]^n\) when n = 0 or a – ve integer

Hence every element \(f\left(a^n\right) \in \mathbf{G}^{\prime}\) can be expressed as \({f(a)}^n\).

∴ f{a) is a generator of G’ implying that G’ is cyclic.

Theorem. 16. Let a he a generator of a cyclic group (G,.) of order n. Then \(a^m\) generates of a cyclic sub-group of (H,.) of (G, .) and O(H)=n/d where d is the H.C.F of n and m.

Proof. am generates a cyclic subgroup (H,.) of (G,.)

Let p be the smallest positive integer such that \((a^m)^p = e\) where e is the identity in H.

Let \(a^m =b\). Let \(b^k \in \mathbf{H} ; k>p\).

Now there exist integers q and r such that k = pq + r,0 ≤ r < p .

∴ \(b^k=b^{p q+r}=b^{p q} \cdot b^r=\left(b^p\right)^q \cdot b^r=e^q \cdot b^r=b^r \text { for } 0 \leq r<p\) .

∴ Any exponent k of b, greater than or equal to p, is reducible to r for 0 ≤ r < p .

∴ H contains p elements given by

∴ \(\mathbf{H} \simeq\left\{b, b^2, \ldots ., b^{p-1}, b^p=e\right\} \text { i.e. } \mathbf{H}=\left\{\left(a^m\right)^1,\left(a^m\right)^2, \ldots .,\left(a^m\right)^p=e\right\}\)

H has p elements, as many elements as the smallest power of \(a^m\) which gives the identity e. Now \(a^pm\) = e if and only if n divides pm since \(a^n\) = e, (G,.) be a cyclic group of order n.

∴ pm/n must be an integer.

Let d be the. H. C. F of n and m. Now \(\frac{p m}{n}=p \cdot \frac{m / d}{n / d}\)

But n/d does not divide m/d.

∴ n/d divides/p. ∴ Least value of p is n/d .

∴ 0 (H) = n/d.

Let | G | = 24 and G be cyclic. If \(a^8 \neq e \text { and } a^{12} \neq e\), show that G = < a > Divisors of 24 are 1,2,3,4,6,8,12, 24.

If | a | = 24 then \(a^2=e \text { and } a^4=\left(a^2\right)^2=e^2=e=a^8\)

Also if |a| = 3 , then \(a^3=e \text { and } a^{12}=\left(a^3\right)^4=e^4=e=a^6\) .

∴ |a|=24 is only acceptable and hence G=<a>.

Theorem 17. A cyclic group of order n has ϕ(n) generators.

Proof. First, we prove Theorem 11.

∴ G = \(<a^m>\) <=> (m,n) = 1 ,

∴ \(<a^m>\) is a generator of G <=> m is a positive integer less than n and relatively prime to n.

⇒ The number of generators of G <=> the number of positive integers that are less than n and relatively prime to n = ϕ(n).

Note. For n = 1, ϕ(1) = 1, and for n > 1 the number of generators ϕ(n) is the number of positive integers less than n and relatively prime to n.

example. a is a generator of a cyclic group G of order 8. Then G = < a > and O (a) = 8 .

Here G = \(\left\{a, a^2, a^3, a^4, a^5, a^6, a^7, a^8\right\}\)

Since 3, 5, 7 are relatively prime to 8 and each is less than \(8, a^3, a^5, a^7\) are the only other generators of G. Also \(a^2, a^4, a^6, a^8\) cannot be the generators of G. Hence G has only 4 generators and they are \(a^1, a^3, a^5,a^7\).

Now < a³ > =\(\left\{a^3, a^6, a^1, a^4, a^7, a^2, a^5, a^8\right\}\), etc.

Cyclic Group Properties And Their Proofs

Example 12. Show that the group (G = \({1,2,3,4,5,6},x_7)\)) is cyclic. Also, write down all its generators.

Solution. Clearly O (G) = 6. If there exists an element a ∈ G such that O (a) = 6, then G will be a cyclic group with generator a.

Since. \(3^1=3,3^2=3 \times_7 3=2,3^3=3^2 \times_7 3=6,3^4=3^3 \times_7 3=4\)

⇒ \(3^5=3^4 \times \times_7=5,3^6=3^5 \times_7 3=1\) , the identity element.

∴ G = \({3, 3^2, 3^3, 3^4, 3^5, 3^6}\) and is cyclic with 3 as a generator.

Since 5 is relatively prime to 6, \(3^5\) i.e. 5 is also a generator of G.

∴ Generators of G are 3 and 5.

Note. If (G, .) is a cyclic group of order n, then the number of generators of G = ϕ(n) = the number of numbers less than n and prime to n.

From theory of numbers, if \(n=p_1^{\alpha_1} \cdot p_2^{\alpha_2} \ldots \ldots p_k^{\alpha_k} \text { where } p_1 \ldots \ldots p_k\) are all primes factors of n, then \(\phi(n)=n\left(1-\frac{1}{p_1}\right) \ldots\left(1-\frac{1}{p_k}\right)\)

⇒ \(\phi(6)=6\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=2\) i.e. G has 2 generators.

Further, if \(n=p^\alpha\)where p is less than and prime to n, then \(\phi(n)=p^\alpha\left(1-\frac{1}{p}\right)\)

Example 13. Find all the subgroups of \(\left(\mathbf{Z}_{18},+_{18}\right)\).

Solution. \(\left(\mathbf{Z}_{18},+_{18}\right)\) is a cyclic group with 1 as its generator.

ο \(Z_18\) = {0,1,2,3, ,1.7} and all subgroups are cyclic.

Now all the generators of the group \(Z_18\) are less than 18 and are prime to 18. Thus 1, 3, 5, 7, 11, 13, and 17 are all generators of \(Z_18\).

All the subgroups of \(Z_18\) are the subgroups generated by 1,2,3,6,9,18 (Divisors of 18).

This number that corresponds to 18 is 0.

The subgroups are:

Trivial (improper)subgroups- \(\left(\mathbf{Z}_{18},+_{18}\right)=\langle 1\rangle,\left(\{0\},+_{18}\right)=\langle 0\rangle\).

proper groups \(\left(\{0,2,4,6,8,10,12,14,16\},+{ }_{18}\right)=\langle 2\rangle\),

⇒ \(\left(\{0,3,6,9,12,15\},++_{18}\right)=\langle 3\rangle\),

⇒  \(\left(\{0,6,12\},+_{18}\right)=\langle 6\rangle,(\{0,9\},+18)=\langle 9\rangle\).

Note.

1.O(<1>)=18, O(<0>)=1,O(<2>)=9,O(<3>)=6,O(<6>)=3,O(<9>)=2.

2. Lattice diagram for \(\left(\mathbf{Z}_{18},+_{18}\right)\).

Example 14. Find the number of elements in the cyclic subgroup of \(\left(\mathbf{Z}_{30},+{ }_{30}\right)\) generated by 25 and hence write the subgroup

Solution. \(\left(\mathbf{Z}_{30},+{ }_{30}\right)\) is a cyclic group.

Clearly 1 is a generator of \(\left(\mathbf{Z}_{30},+{ }_{30}\right)\).

Now \(25 \in \mathbf{Z}_{30} \text { and } 25=1^{25}=(25)(1)\).

Clearly \(\left(1^{25},+_{30}\right)\) is a subgroup of \(\left(\mathbf{Z}_{30},+{ }_{30}\right)\)

The g.c.d. of 30 and 25 is 5.

∴ \(25=1^{25}\) generates a cyclic subgroup of order (30/5) = 6

i.e. ({0,5,10,15,20,25},+30) is the cyclic subgroup generated by 25.

Example 15. Find the no. of elements in the cyclic subgroup of \(\left(\mathbf{Z}_{42},+_{42}\right)\) generated by 30 and hence write the subgroup.

Solution. \(\left(\mathbf{Z}_{42},+_{42}\right)\) is a cyclic group.

Clearly 1 is a generator of \(\left(\mathbf{Z}_{42},+_{42}\right)\)

Now \(30 \in \mathbf{Z}_{42} \text { and } 30=1^{30}=(30) \text { (1) }\) .

Clearly \(\left(1^{30},+_{42}\right)\) is a subgroup of \(\left(\mathbf{Z}_{42},+_{42}\right)\) .

The g.c.d. of 30 and 42 is 6.

∴ \(30=1^{30}\) generates a cyclic subgroup of order (42/6) = 7

i.e.\(\left(\{0,6,12,18,24,30,36\},+_{42}\right)\) is the cyclic subgroup generated by 30.

Example 16. Find the order of the cyclic subgroup of \(\left(\mathbf{Z}_{60},+_{60}\right)\) generated by 30.

Solution. \(\left(\mathbf{Z}_{60},+_{60}\right)\) is a cyclic group and 1 is a generator of it.

Now \( 30 \in \mathbf{Z}_{60} \text { and } 30=1^{30}=30 \text { (1) }\).

Clearly \(\left(1^{30},+_{60}\right) \text { i.e. }\left(30,++_{60}\right)\) is a subgroup of \(\left(\mathbf{Z}_{60},+_{60}\right)\) .

The g.c.d. of 60 and 30 is 30.

∴ \(30 = l^30\) generates a cyclic subgroup of order (60/30) = 2 .

Example 17. Find the number of generators of cyclic groups of orders 5,6,8,12,15,60.

Solution: O(G)=5, the number of generators of \(\mathbf{G}=\phi(5)=5\left(1-\frac{1}{5}\right)=4\)

O(G)=6, the number of generators of G = ϕ(6)

=\(6\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=2\) (∵ 2,3 are prime factors of 6)

O(G)=8, the number of generators of G

=\(\phi(8)=8\left(1-\frac{1}{2}\right)=4\) (∵ 2 is the only prime factor of 8)

O(G)=12, the number of generators of G

=\(\phi(12)=12\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=4\) (∵ 2,3 are the only prime factors of 12)

O(G)=15, the number of generators of G = \(\phi(15)=15\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)=8\)

(∵ 3,5 are the only prime factors of 15.)

O(G)=60, the number of generators of

= \(\mathbf{G}=\phi(60)=60\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)=16\).

(∵ \(60=2^2 \cdot 3 \cdot 5 ; 2,3,5\) are the only prime factors of 60)

Example 18. Show that \(\left(\mathbf{Z}_p,+_p\right)\) has no proper subgroups if p is prime.

Solution. \(\left(\mathbf{Z}_p,+_p\right)\) is a cyclic group and \(\mathbf{O}\left(\mathbf{Z}_p\right)\) = p where p is prime.

∴ Number of generators of \(\left(\mathbf{Z}_p,+_p\right)=p\left(1-\frac{1}{p}\right)\)

∴ All the p -1 elements of \(Z_p\) except the identity element, generate the group \(\left(\mathbf{Z}_p,+_p\right)\). But this is a trivial subgroup.

So \(\left(\mathbf{Z}_p,+_p\right)\) has no proper subgroups.

Example 19. Find all orders of subgroups of the group \(\mathbf{z}_{17}\).

Solution. \(\left(\mathbf{Z}_{17},+_{17}\right)\) is a cyclic group and \(\mathbf{O}\left(\mathbf{Z}_{17}\right)=17\) (17 is prime)

∴ The no. of generators of \(\left(\mathbf{Z}_{17},+_{17}\right)=17\left(1-\frac{1}{17}\right)=16\)

The 16 elements of \(\mathbf{z}_{17}\) except identity element 0 (which corresponds to 17), generate the group \(\left(\mathbf{Z}_{17},+_{17}\right)\) which is of course a trivial subgroup.

Hence \(\left(\mathbf{Z}_{17},+_{17}\right)\) has no proper subgroups.

Also {0} is a trivial subgroup. Now \(\mathbf{O}\left(\mathbf{Z}_{17}\right)=17, \mathbf{O}(\{0\})=1\).

Example 20. (1) If p, q be prime numbers, find the number of generators, of the cyclic group \(\left(\mathbf{Z}_{p q},+_{p q}\right)\).

Solution.

Given

p, q be prime numbers

The number of generators of \(\left(\mathbf{Z}_{p q},+_{p q}\right)\)

=\(\phi(p q)=p q\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\)(∵ p,q are prime)

(2) If p is a prime number, find the number of generators of the cyclic group \(\left(\mathbf{Z}_{p^r},+_{p^r}\right)\) where r is an integer ≥ 1.

Solution. The number of generators = \(\phi\left(p^r\right)=p^r\left(1-\frac{1}{p}\right)=p^{r-1}(p-1)\)

Example 21. G is a group. If a is the only element in G such that \(|\langle a\rangle|=2\), then show that for every x ∈ G, ax = xa.

Solution.

Given

G is a group. If a is the only element in G such that \(|\langle a\rangle|=2\)

a is the only element in group G such that | < a > | = 2.

Let e be the identity in G.

∴ \(a^2=e\).

Also whenever\(b \in \mathbf{G} \Rightarrow b^2=e\), we have b = a

Now for \(x \in \mathbf{G}, x a x^{-1} \in \mathbf{G}\).

∴ In G , \(\left(x a x^{-1}\right)^2=\left(x a x^{-1}\right)\left(x a x^{-1}\right)=x a\left(x^{-1} x\right) a x^{-1}=x \text { e } a x^{-1}\)

=\(x \text { a } a x^{-1}=x a^2 x^{-1}=x \text { e } x^{-1}=x x^{-1}=e\).

∴ \(x a x^{-1}=a \Rightarrow x a x^{-1} x=a x \Rightarrow x a e=a x \Rightarrow x a=a x\).

Example. 22. Find all cosets of the subgroup < 4 > of \(\mathbf{Z}_{12}\)

Solution. \(\left(\mathbf{Z}_{12}=\{0,1,2, \ldots ., 11\},+_{12}\right)\)is a cyclic group.

Let the subgroup < 4 > of \(\mathbf{z}_{12}\) be H.

Since <4> = { ,-12,-8,-4,0,4,8,12,….}, \(\left(\mathbf{H}=\{0,4,8\},+_{12}\right)\) is the subgroup of \(\left(\mathbf{Z}_{12},+_{12}\right)\) for which all left cosets have to be found out.

Composition Table:

⇒ \(0+_{12} \mathbf{H}=\{0,4,8\}, 1+{ }_{12} \mathbf{H}=\{1,5,9\}\)

⇒ \(2+_{12} \mathbf{H}=\{2,6,10\}, 3+_{12} \mathbf{H}=\{3,7,11\}\)

⇒ \(4+_{12} \mathbf{H}=\{4,8,0\}, 5+_{12} \mathbf{H}=\{5,9,1\}\)

⇒ \(6+_{12} \mathbf{H}=\{6,10,2\}, 7+_{12} \mathbf{H}=\{7,11,3\}\)

………………………………………

⇒ \(11+_{12} \mathbf{H}=\{11,3,7\}\)

∴ \(0+_{12} \mathbf{H}=4+_{12} \mathbf{H}=\ldots . .=\{0,4,8\}\)

⇒ \(1+_{12} \mathbf{H}=5+_{12} \mathbf{H}=\ldots \ldots=\{1,5,9\}\)

⇒ \(2+_{12} \mathbf{H}=6+_{12} \mathbf{H}=\ldots . .=\{2,6,10\}\)

⇒ \(3+_{12} \mathbf{H}=7+_{12} \mathbf{H}=\ldots \ldots=\{3,7,11\}\)

Since \(\left(\mathbf{Z}_{12},+_{12}\right)\) is abelian, left cosets of H are also right cosets of H.

In \(\mathbf{Z}_{12}\) cosets of H are \(0+_{12} \mathbf{H}, 1++_{12} \mathbf{H}, \ldots ., 11+_{12} \mathbf{H} \text { or } \mathbf{H}++_{12} 0, \mathbf{H}+{ }_{12} 1, \ldots ., \mathbf{H}++_{12} 11\)

Also \(0+_{12} \mathbf{H}=\mathbf{H}, 1+_{12} \mathbf{H}, 2+_{12} \mathbf{H}, 3+_{12} \mathbf{H}\)

Example 23. Find all cosets of the subgroup <18> of \(\mathbf{Z}_{36}\)

Solution. \(\left(\mathbf{Z}_{36}=\{0,1,2,3, \ldots \ldots, 35\},+_{36}\right)\) is a finite cyclic abelian group.

The subgroup < 18 > of \(\mathbf{Z}_{36}\) is cyclic and let it be denoted by H.

∴ H = {0,18}.

Here + means +36.

∴ Left cosets of H in \(\mathbf{Z}_{36}\) are

0+ H = {0,18} 18 + H = (18,0}

1+ H = {1,19} 19 +H = {19,1}

2+ H = {2,20} 20 + H = {20,2}

17+ H = {17,35} 35 + H = {35,17}

∴ Distinct left cosets of H in \(\mathbf{Z}_{36}\) are 0+ H,1 + H,….,17+ H and their number is 18

Since\(\mathbf{G}=\left(\mathbf{Z}_{36},+_{36}\right)\) is abelian,

left coset of H in G = right coset of H in G.

Cosets of <18> of \(\mathbf{Z}_{36}\) are 0 + H,1 + H,….,17 + H dr H + 0,H + 1,….,H + 17 .

Example 24. \(S_5\) is the set of all permutations on 5 symbols is a group. Find the index of the cyclic subgroup generated by the permutation (1 2 4) in \(S_5\)

Solution: Let f = (1 2 4).

∴ \(f=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 3 & 1 & 5
\end{array}\right)\)

∴ \(f^2=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 3 & 1 & 5
\end{array}\right)\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 3 & 1 & 5
\end{array}\right)=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
4 & 1 & 3 & 2 & 5
\end{array}\right)\) and

∴ |<f>|=3 and |\(S_5\)| = 5! = 120.

∴ Index of the cyclic subgroup f in \(\mathrm{S}_5=\frac{\left|\mathrm{S}_5\right|}{|\langle f\rangle|}=\frac{120}{3}=40\)

Example 25. If \(\sigma=\left(\begin{array}{llll}
1 & 2 & 3 & 4 \\
2 & 1 & 3 & 4
\end{array}\right), \tau=\left(\begin{array}{llll}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1
\end{array}\right)\) are two permutations defind on A={1,2,3,4}, find the cyclic groups generated by σ,τ.

Solution: If n is a least positive integer such that \(f^n=e\) where f is a permutation on A,

then <f>=\(\left\{\mathrm{I}, f, f^2, \ldots \ldots, f^{n-1}\right\}\)

Now \(\sigma^2=\left(\begin{array}{llll}
1 & 2 & 3 & 4 \\
2 & 1 & 3 & 4
\end{array}\right)\left(\begin{array}{llll}
1 & 2 & 3 & 4 \\
2 & 1 & 3 & 4
\end{array}\right)=\left(\begin{array}{llll}
1 & 2 & 3 & 4 \\
1 & 2 & 3 & 4
\end{array}\right)=I \Rightarrow\langle\sigma\rangle=\{I, \sigma\}\)

Also \(\tau^2=\left(\begin{array}{llll}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1
\end{array}\right)\left(\begin{array}{llll}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1
\end{array}\right)=\left(\begin{array}{llll}
1 & 2 & 3 & 4 \\
3 & 4 & 1 & 2
\end{array}\right)\)

⇒ \(\langle\tau\rangle=\left\{\mathrm{I}, \tau, \tau^2, \tau^3\right\}\)

Example 26. If \(\mathrm{S}=\left\{\mathrm{P}, \mathrm{P}^2, \mathrm{P}^3, \mathrm{P}^4, \mathrm{P}^5, \mathrm{P}^6\right\} \text { with } \mathrm{P}=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 5 & 1 & 3
\end{array}\right)\) then by using a multiplication table, prove that S forms an abelian group.

Solution. \(\mathrm{S}=\left\{\mathrm{P}, \mathrm{P}^2, \mathrm{P}^3, \mathrm{P}^4, \mathrm{P}^5, \mathrm{P}^6\right\} \text { with } \mathrm{P}=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 5 & 1 & 3
\end{array}\right)\)

Permutation multiplication is the operation.

⇒ \(P^2=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 5 & 1 & 3
\end{array}\right)\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 5 & 1 & 3
\end{array}\right)=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
4 & 1 & 3 & 2 & 5
\end{array}\right)\)

⇒ \(P^3=P^2 P=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
4 & 1 & 3 & 2 & 5
\end{array}\right)\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 5 & 1 & 3
\end{array}\right)=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 5 & 4 & 3
\end{array}\right)\)

⇒ \(=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 3 & 1 & 5
\end{array}\right)\)

⇒ \(P^5=P^4 P=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 3 & 1 & 5
\end{array}\right)\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 5 & 1 & 3
\end{array}\right)\)

= \(\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
4 & 1 & 5 & 2 & 3
\end{array}\right)\)

⇒ \(P^6=P^5 P=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
4 & 1 & 5 & 2 & 3
\end{array}\right)\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 5 & 1 & 3
\end{array}\right)=\left(\begin{array}{lllll}
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 3 & 4 & 5
\end{array}\right)=I\)

Clearly, \(P^2 P^3=P^3 P^2, P^3 P^4=P^4 P^3\)

Here \(P^6=I, \quad P^{-1}=P^5, P^5=P^{-1}\)etc. Thus S is an abelian group.

Example 27. Let A = {1,2,3}. Find the cyclic subgroups generated by

⇒ \(P_1=\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right), P_2=\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right), M_1=\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 3 & 2
\end{array}\right) \text { of } S_3\)

Solution: Let A = {1,2,3}. ∴ \(S_3=\left\{e, P_1, P_2, M_1, M_2, M_3\right\}\)

where \(e=\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 2 & 3
\end{array}\right)\)

⇒ \(P_1=\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right), P_2=\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right), M_1=\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 3 & 2
\end{array}\right),\)

⇒ \(M_2=\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 3
\end{array}\right), M_3=\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right)\)(say)

⇒ \(P_1^2=\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
⇒ \end{array}\right)\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right)=\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right)=P_2\)

∴ \(<P_1>=\left\{e, P_1, P_2\right\}\)

⇒ \(P_2^2=\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right)\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right)=\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right)=P_1\)

⇒ \(P_2^3=P_2^2 P_2=\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right)\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right)=\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 2 & 3
\end{array}\right)=e\)

∴\(\left.<P_2\right\rangle=\left\{e, P_1, P_2\right\}\)

⇒ \(\mathrm{M}_1^2=\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 3 & 2
\end{array}\right)\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 3 & 2
\end{array}\right)=\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 2 & 3
\end{array}\right)=e\)

∴ \(<\mathrm{M}_1>=\left\{e, \mathrm{M}_1\right\}\)

Example 28. If \(f: \mathbf{G} \rightarrow \mathbf{G}^{\prime}\) is isomorphic then the order of an element in G is equal to the order of its image in G’.

Solution. Since f is one – one onto mapping, corresponding to any element a’ ∈ G’ there exists an element a ∈ G. such that f{a) = a’. If e is the identity in G and e’ is the identity in G’ we have f(e) = e’.

Let n be the order of a ∈ G so that \(a^n = e\) where n is the least positive integer.

We have to show that the order of the image f{a) of a is also n.

Now \(a^n=e \Rightarrow f\left(a^n\right)=f(e)=e^{\prime}\)

⇒ f{a .a .a…. n times) = e’ ⇒ f(a).f(a). f (a)…… to n times = e’

⇒ [f(a)]n = e’ => the order of \(f(a) \leq n\) .

Let us suppose that m is the order of f{a) where m < n

so that \([f(a)]^m=e^{\prime}=f(e)\)

i.e. f(a). f(a). f(a)…. m times = f(e).

i.e. f(a. a . a…. m times) = f(e) i.e. \(f\left(a^m\right)=f(e) \Rightarrow a^m=e\)

Since m is less than n, \(a^m = e\) is a contradiction.

Hence there cannot be any other integer m less than n such that \(a^m = e\) .

∴ m = n ⇒ Order of f(a) = Order of a

⇒ Order of the image of an element = order of that element.

Theorem 18. If G is an infinite cyclic group, then G has exactly two generators which are inverses of each other.

Proof. Let G be an infinite cyclic group generated by a.

∴ \(\mathbf{G}=\left\{a^n / n \in \mathbf{Z}\right\}\).

Let \(a^m\) be a generator of G.

Since \(a \in \mathbf{G}, \exists\) an integer p such that \(a=\left(a^m\right)^p\).

i.e. \(a^{m p}=a\)

i.e.\(a^{m p} a^{-1}=a \cdot a^{-1}\)

i.e.\(a^{m p-1}=e\)

If mp – 1 > 0 then ∃q = mp – 1 such that \(a^q = e\) implies that G is finite.

But G is infinite.

∴ mp-1=0

i.e. mp = li.e. m = ±1,p = ±1 ,

∴ \(a^1, a^{-1}\) are generators of G.

i.e. G has exactly two generators and one is the inverse of the other in G.

Note. (Z, +) is an infinite cyclic group and it has only two generators 1 and – 1.

Theorem 19. Any infinite cyclic group is isomorphic to the additive group of integers (Z, +)

Proof: Let G be an infinite cyclic group generated by an element \(\mathbf{a}(\in \mathbf{G})\)

Thus 0(a) = 0 or ∞ and \(a^0 = e\) (identity in G)

∴ \(G = \left\{a^n / n \in \mathbf{Z}\right\}\) and all the elements of G are distinct.

Define a mapping \(f: \mathbf{G} \rightarrow \mathbf{Z} \text { such that } f\left(a^n\right)=n, \forall a^n \in \mathbf{G}\)

Let \(a^i, a^j \in \mathbf{G}\) . Let (Z, +) be the additive group of integers.

Now \(f\left(a^i\right)=f\left(a^j\right) \Rightarrow i=j \Rightarrow a^i=a^j\)

∴ f is 1 -1.

Let \(k \in \mathbf{Z}\)

∴ \(a^k \in \mathbf{G} \text { and } f\left(a^k\right)=k\)

∴ f is onto.

Further \(a^i, a^j \in \mathbf{G}\)and\(f\left(a^i a^j\right)=f\left(a^{i+j)}=i+j\right)\)\(f\left(a^i\right)+f\left(a^j\right)\).

∴ f is a homomorphism and hence/is an isomorphism from G to Z.

∴ \(G \cong Z\)

Theorem 20: Every finite cyclic group G of order u is isomorphic to the group of integers addition modulo n, i.e.\(\left(Z_n+{ }_n\right)\)

Proof: Let G be a finite cyclic group of order n generated by an element \( a(\in \mathbf{G})\).

Let e be the identity in G.

∴ \(G = \left\{a^0=e, a, a^2, a^3, \ldots \ldots . ., a^{n-1}\right\}=\left\{a^m / m \text { is an integer and } 0 \leq m<n\right\}\)

⇒ \(\mathbf{Z}_n=\{0,1,2, \ldots . .(n-1)\}\) is the group of integers w.r.t. +n.

Define a mapping \(f: \mathbf{G} \rightarrow \mathbf{Z}_n \text { such that } f\left(a^m\right)=m \forall a^m \in \mathbf{G}\) .

Since \(a^0=e, f(e)=f\left(a^0\right)=0\) where 0 s the identity in \(\left(\mathrm{Z}_n+{ }_n\right)\)

Let \(a^i, a^j \in \mathbf{G}\).

Now \(f\left(a^i\right)=f\left(a^j\right) \Rightarrow i=j \Rightarrow a^i=a^j\)

∴ f is 1 -1.

Let \(k \in \mathbf{Z}_n\)

⇒ \(a^k \in \mathbf{G} \text { and } f\left(a^k\right)=k\)

∴ f is onto.

Let \(a^i, a^j \in \mathbf{G}\). Then \(a^i, a^j \in \mathbf{G}\) and \( f\left(a^i a^j\right)=f\left(a^{l+j}\right)\). By division algorithm, there exist integers q and r.

such that i + j = qn + r, 0 ≤ r < n.

∴ \(a^{i+j}=a^{q n+r}=\left(a^n\right)^q \cdot a^r=e^q a^r=a^r\)

(∵\(a^n=a^0=e\))

∴ \(f\left(a^i a^j\right)=f\left(a^{i+j}\right)=f\left(a^r\right)=r\)

∴ \(f\left(a^i\right)++_n f\left(a^j\right)=r\) by the definition of f.

∴ f is a homomorphism and hence f is an isomorphism from G to \(\mathbf{Z}_n\).

∴ \(\mathbf{G} \cong \mathbf{Z}_n\)

Coordinates Of A Point In Space

Let O be any point in space S. Let \(\overleftrightarrow{X^{\prime} X}, \overleftrightarrow{Y^{\prime} Y}\) be two perpendicular lines through O. Let the plane determined by the lines be \(\overleftrightarrow{\mathrm{XOY}}\).

Through O and perpendicular to the plane \(\overleftrightarrow{\mathrm{XOY}}\) let \(\overleftrightarrow{Z^{\prime} Z}\) be a line.

Imagine the plane \(\overleftrightarrow{\mathrm{XOY}}\) as the plane of the paper and the perpendicular \(\overleftrightarrow{Z^{\prime} Z}\) is to be visualized as perpendicular lines through O.

⇒ \(\overleftrightarrow{Z^{\prime} Z}\) may be regarded as vertical and \(\overleftrightarrow{X^{\prime} X}, \overleftrightarrow{Y^{\prime} Y}\) as horizontal. ⇒ \(\overleftrightarrow{X^{\prime} X}, \overleftrightarrow{Y^{\prime} Y}, \overleftrightarrow{Z^{\prime} Z}\) are three non-coplaner mutual perpendicular lines through O.

The lines \(\overleftrightarrow{Y^{\prime} Y}\) and \(\overleftrightarrow{Z^{\prime} Z}\) determine the plane \(\overleftrightarrow{\mathrm{YOZ}}\) and the lines \(\overleftrightarrow{Z^{\prime} Z}, \overleftrightarrow{X^{\prime} X}\) determine the plane \(\overleftrightarrow{\mathrm{ZOX}}\).

Also \(\overleftrightarrow{\mathrm{XOY}},\overleftrightarrow{\mathrm{YOZ}},\overleftrightarrow{\mathrm{ZOX}}\) are three mutually perpendicular planes through O.

On \(\overleftrightarrow{X^{\prime} X}\) take O as the origin, I as the unit point;

On \(\overleftrightarrow{Y^{\prime} Y}\) take O as the origin, J as the unit point and

On \(\overleftrightarrow{Z^{\prime} Z}\) take O as the origin, K as the unit point such that \(\overline{\mathrm{OI}}=\overline{\mathrm{OJ}}=\overline{\mathrm{OK}}\)

The coordinate of a point on \(\overleftrightarrow{X^{\prime} X}\) is called its x-coordinate, the coordinate of a point on \(\overleftrightarrow{Y^{\prime} Y}\) is called its y-coordinate and the coordinate of a point on \(\overleftrightarrow{Z^{\prime} Z}\) is called its z-coordinate.

⇒ \(\overleftrightarrow{\mathrm{XOY}}\)(XY plane), \(\overleftrightarrow{\mathrm{YOZ}}\)(YZ plane),\(\overleftrightarrow{\mathrm{ZOX}}\)(ZX plane) are called the rectangular coordinate planes.\(\overleftrightarrow{X^{\prime} X}\)(x-axis), \(\overleftrightarrow{Y^{\prime} Y}\)(y-axis),\(\overleftrightarrow{Z^{\prime} Z}\)(z-axis) are called the rectangular coordinate axes.

Such an assigned system of axes is called the frame of reference or coordinate frame (denoted by OXYZ) so that the coordinates of a point change with the change in the frame of reference. P is any point in space.

L, M, and N are respectively the projections on the coordinates axes. Let the X coordinate of L be x, Y coordinate of M be y and Z coordinate of N be z. The numbers x, y, z taken in this order are called the rectangular coordinates of P. We write p = (x, y, z). Thus point P is associated with an ordered trias of real numbers.

Let (x, y, z) be an ordered triad. Take the point L of co-ordinate x on \(\overleftrightarrow{X^{\prime} X}\), the point M of coordinate y on \(\overleftrightarrow{Y^{\prime} Y}\) and the point N of coordinate z on \(\overleftrightarrow{Z^{\prime} Z}\).

Through the points L, M, N draw planes π₁, π₂, π₃ perpendicular to the coordinate axes \(\overleftrightarrow{X^{\prime} X}\),\(\overleftrightarrow{Y^{\prime} Y}\),\(\overleftrightarrow{Z^{\prime} Z}\) respectively. The three mutually perpendicular planes π₁, π₂, π₃ intersect at a unique point P.

Since P ∈ π1, the projection of P on \(\overleftrightarrow{X^{\prime} X}\) is L and hence the x-coordinate of p is equal to L. Similarly the y-coordinate of p is equal to the y-coordinate of M and the z-coordinate of p is equal to the z-coordinate of N.

The coordinates of P are x, y, and z in that order. Thus for the ordered triad (x, y, z), we have a unique point P in space.

Hence a one-to-one correspondence is established between the set of points in space and the set of ordered traids of real numbers. This space is called 3D space or R₃ space.

The three coordinate planes divide the space into eight compartments, each of which is called an octant.

Clearly, the three planes through P are respectively parallel to the coordinates planes and the six planes form a rectangular parallelopiped. The three pairs of rectangular faces are

Coordinates Definition And Solved Exercise Problems

PFNG, EMOL; PGLE, FNOM; PEMF, GLON

We have

(1) For any point on the X-axis, y = 0 and z = 0.

For any point on the Y-axis, x = 0 and z = 0.

For any point on the Z-axis, x = 0 and y = 0.

(2) For any point on \(\overleftrightarrow{\mathrm{XOY}}\) plane z = 0.

For any point on \(\overleftrightarrow{\mathrm{YOZ}}\) plane x = 0.

For any point on \(\overleftrightarrow{\mathrm{ZOX}}\) plane y = 0.

(3) Origin O = (0, 0, 0)

(4) |x| = OL = ME = NG = FP = Distance of p from YZ plane,

|y| = OM = LE = NF = GP = Distance of p fron ZX plane,

|z| = ON = LG = MF = EP = Distance of p from XY plane.

(5) PL, PM, and PN are respectively perpendicular to X axis, Y axis, Z axis.

(6) Distance of p from the X-axis = PL = \(\sqrt{\left(\mathrm{LE}^2+\mathrm{EP}^2\right)}=\sqrt{\left(y^2+z^2\right)}\)

Distance of P from the Y-axis = PM = \(\sqrt{\left[\mathrm{EP}^2+\mathrm{ME}^2\right]}=\sqrt{\left(z^2+x^2\right)}\), (∵ (\(\overrightarrow{\mathrm{EP}}, \overrightarrow{\mathrm{EM}})=90^{\circ}\))

Distance of P from the Z-axis = PN = \(\sqrt{\left[\mathrm{NF}^2+\mathrm{FP}^2\right]}=\sqrt{\left(x^2+y^2\right)}\), (∵ (\(\overrightarrow{\mathrm{FP}}, \overrightarrow{\mathrm{FN}})=90^{\circ}\))

(7) \(\mathrm{OP}=\sqrt{\left[\mathrm{OL}^2+\mathrm{PL}^2\right]}=\sqrt{\left(x^2+y^2+z^2\right)}\) (∵ (\(\overrightarrow{\mathrm{LO}}, \overrightarrow{\mathrm{LP}})=90^{\circ}\))

(8) The projection of p(x, y, z) on the coordinates axes are L, M, N where L = (x, 0, 0), M = (0, y, 0), and N = (0, 0, z).

The projections of p(x, y, z) on the coordinate planes (YZ, ZX, XY) are F, G, E. where F = (0, y, 0), G = (x, 0 z), E = (x, y, 0).

(9) \(\begin{array}{ll}
\mathrm{X} \text { axis }=\{\mathrm{P}(x, y, z) / y=0, z=0\}, & \mathrm{YZ} \text { plane }=\{\mathrm{P}(x, y, z) / x=0\}, \\
\mathrm{Y} \text { axis }=\{\mathrm{P}(x, y, z) / x=0, z=0\}, & \mathrm{XY} \text { plane }=\{\mathrm{P}(x, y, z) / z=0\}, \\
\text { Z axis }=\{\mathrm{P}(x, y, z) / x=0, y=0\}, & \text { ZX plane }=\{\mathrm{P}(x, y, z) / y=0\} .
\end{array}\)

Coordinates Interpretation Of Equations

Definition. A locus is the set of points and only those points satisfying a given condition.

Definition. F is a function from R3 into R.

Then the locus S = {(x, y, z)|F(x, y, z) = 0} is called the surface represented by the equation F(x, y, z) = 0.

F(x, y, z) = 0 is called an equation to the surface S.

Definition. If F is a polynomial (not a zero polynomial) in x, y, z then the locus (surface) represented by F(x, y, z) = 0 is called an algebraic surface.

Consider.

(1) F(x)=0 where F(x) is a polynomial. So F(x) = 0 has a finite number of real roots. Let x₁,x₂,…..,x_n be the real roots. The locus (surface) \(\pi_1=\left\{(x, y, z) / x=x_1\right\}\) is a plane parallel to YZ plane since every point in π₁ has its x coordinate x₁

Similarly, the locus (surface) of each of the equations x = x₂,….,x = xn is a plane parallel to YZ plane. Thus the locus of the equation F(x) = 0 is a system of planes parallel to the YZ plane.

Similarly, F(y) = 0 and F(z) = 0 can be interpreted.

(2) F(x,z) = 0 where is a first degree polynomial in x, z (say, 3x – 5z = 15). This is a line in ZX plane. The locus (surface) π = {(x, y, z)|F(x, z) = 0} is a plane parallel to the Y axis since every point P(x, y, z) in π has a point on the line with the same x and z coordinates.

Similarly, F(x, y) = 0 where F(x, y) is a first-degree polynomial is the equation of a plane parallel to Z axis and F(y, z) = 0 where F(y, z) is a first-degree polynomial is the equation of a plane parallel to the x-axis.

(3) The curve F(x, z) = 0 is of second degree in the ZX plane in the two-dimensional Cartesian system.

Let Q(x, 0, z) be any point on F(x, z) = 0. Then the point P(x, y, z) where y ∈ R satisfies the equation F(x, z) = 0. \(\overleftrightarrow{\mathrm{PQ}}\) is a line parallel to Y-axis.

∴ For all Q on F(x, z) = 0, we have a set of lines parallel to the y-axis.

Thus the locus of the equation F(x, z) = 0 in 3D space is a system of lines parallel to the y-axis and the locus is a surface called a cylinder. Each of the lines is called a generator of the cylinder.

Similarly, F(x, z) = 0 and F(z, x) = 0 can be interpreted.

A surface generated by a line so that it keeps parallel to a fixed line and intersects a fixed curve in a plane is called a cylindrical surface or a cylinder. In accordance with this definition, a plane in (1) , or (2) is a special case of a cylinder.

Thus the locus (surface) or an equation in two variables is a cylinder with generators parallel to the axis of the missing variable.

(4) If F(x, y, z) = 0 and Φ (x, y, z) = 0 separately represent two surfaces then the points satisfying both equations lie on the curve of the intersection of the two surfaces.

The study of 3D geometry is made through vector methods wherever feasible. The students are already familiar with a detailed study of the geometrical concept of vectors in Vector Algebra. By expressing the equivalence of a vector to an ordered triad, we recapitulate the necessary ideas of Vector Algebra in the ensuing articles.

If felt convenient, methods using concepts of Vector Algebra may be used by students while proving theorems or solving problems.

Coordinates Vector

Let OXYZ be a frame of reference and \(\bar{i}, \bar{j}, \bar{k}\) be a unit orthogonal vector ordered triad (along \(\overline{\mathrm{OX}}, \overline{\mathrm{OY}}, \overline{\mathrm{OZ}}\)) in the right-handed system.

Let P(x, y, z) be any point in space and be determined by its position vector \(\overline{\mathrm{OP}}\). Then we can have \(\overline{\mathrm{OP}}=x \bar{i}+y \bar{j}+z \bar{k}\) for unique scalars x, y, z.

In view of this, let the point (x, y, z) be associated with a unique vector \(x \bar{i}+y \bar{j}+z \bar{k}\) and the vector \(x \bar{i}+y \bar{j}+z \bar{k}\) be associated with a unique point (x, y, z).

Thus in R³-space, a one-to-one correspondence is established between the set of points and the set of position vectors. Hence we write (x, y, z) = \(x \bar{i}+y \bar{j}+z \bar{k}\)

The coordinate of any point P are the rectangular components of its position vector \(\overline{\mathrm{OP}}\).

Any point on x-axis = (x, 0, 0) = \(x \bar{i}\), any point on the y-axis (0, y, 0) = \(y \bar{j}\) and any point on the z-axis (0, 0, z) = \(z \bar{k}\)

If \(\bar{a}\) istaken to represent the position vector of the point A(x, y, z), then

A \(=\bar{a}=x_1 \bar{i}+y_1 \bar{j}+z_1 \bar{k}=\left(x_1, y_1, z_1\right)\)

Theorems On Plane Coordinates With Examples

Note. \(\overline{\mathrm{O}}=0 \bar{i}+0 \bar{j}+0 \bar{k}=(0,0,0)\).

⇒ \(\overline{\mathrm{P}}=\left(x_1, y_1, z_1\right), \overline{\mathrm{Q}}=\left(x_2, y_2, z_2\right)\) are any two points. Then

(1) \(\overline{\mathrm{OP}}+\overline{\mathrm{OQ}}=\left(x_1 \bar{i}+y_1 \bar{j}+z_1 \bar{k}\right)+\left(x_2 \bar{i}+y_2 \bar{j}+z_2 \bar{k}\right)\)

= \(\left(x_1+x_2\right) \bar{i}+\left(y_1+y_2\right) \bar{j}+\left(z_1+z_2\right) \bar{k}\)

= \(\left(x_1+x_2, y_1+y_2, z_1+z_2\right)\)

(2) \(\overline{\mathrm{OP}}-\overline{\mathrm{OQ}}=\left(x_1-x_2, y_1-y_2, z_1-z_2\right)\).

(3) \(\lambda \overline{\mathrm{OP}}=(\lambda(x \bar{i}+y \bar{j}+z \bar{k})=\lambda x \bar{i}+\lambda y \bar{j}+\lambda z \bar{k}=(\lambda x, \lambda y, \lambda z)\) where λ is a real number.

(4) \(\overline{\mathrm{PQ}}=\overline{\mathrm{PO}}+\overline{\mathrm{OQ}}=\overline{\mathrm{OQ}}-\overline{\mathrm{OP}}=\left(x_2-x_1, y_2-y_1, z_2-z_1\right)\)

Definition. If \(\bar{a}=\overline{\mathrm{AB}}, \bar{b}=\overline{\mathrm{CD}}\) such that A, B, C, D are collinear or \(\overleftrightarrow{A B} \| \overleftrightarrow{C D}\), then \(\bar{a}, \bar{b}\) are said to be parallel or collinear.

Sometimes we write \(\bar{a} \| \overleftrightarrow{\mathrm{CD}} \text { or } \overleftrightarrow{\mathrm{AB}} \| \vec{b}\).

If \(\bar{a}, \bar{b}\) are not parallel or not collinear, then \(\bar{a}, \bar{b}\) are called non-parallel or non-collinear vectors.

P \(=\left(x_1, y_1, z_1\right), \mathrm{Q}=\left(x_2, y_2, z_2\right)\) are any two points. O, P, and Q are collinear

<=> \(\overline{\mathrm{OP}}=\lambda \overline{\mathrm{OQ}}, \lambda\) is a real number

<=> \(\left(x_1, y_1, z_1\right)=\lambda\left(x_2, y_2, z_2\right) \Leftrightarrow x_1: x_2=y_1: y_2: z_2=\lambda: 1\)

Coordinates Length Or Magnitude Of A Vector

Any point \(\overline{\mathrm{P}}=(x, y, z) \text { and } \overline{\mathrm{OP}}=(x, y, z)\). The length or magnitude or norm or modulus of the vector \(\overline{\mathrm{OP}}=|\overline{\mathrm{OP}}|=\overline{\mathrm{OP}}=\sqrt{\left(x^2, y^2, z^2\right)}\)

Theorem 1. Distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is \(\sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2\right]}\).

Proof. Let A = (x₁, y₁, z₁), B = (x₂, y₂, z₂).

Complete the parallelogram OABP so that OP ∥ AB and OP = AB.

∴ \(\overline{\mathrm{OP}}=\overline{\mathrm{AB}}\)

∴\(\overline{\mathrm{OP}}=\left(x_2-x_1, y_2-y_1, z_2-z_1\right)\)

⇒ \(|\overline{O P}|=O P=\sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2\right]}\)

∴ \(|\overline{\mathrm{OP}}|=\mathrm{OP}=\sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2\right]}\)

∴ \(|\overline{\mathrm{AB}}|=\mathrm{AB}=\sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2\right]}\)

∴ Distance between points A = (x₁, y₁, z₁), B = (x₂, y₂, z₂)

= \(\mathrm{AB}=\sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2\right]}\)

If \(\bar{a}, \bar{b}\) are two vectors, then there exists a unique vector \(\bar{c}\) such that \(\bar{c}+\bar{b}=\bar{a}\) i.e., \(\bar{a}\) = (x₁, y₁, z₁), \(\bar{b}\) = (x₂, y₂, z₂), \(\bar{c}\) = (x, y, z) then (x, y, z) + (x₂, y₂, z₂) = (x₁, y₁, z₁) ⇒ (x₁, y₁, z₁) = (x₁ – x₂, y₁ – y₂, z₁ – z₂)

Coordinates Unit Vector

If A, B, and A ≠ B, are points, then \(\frac{\overline{\mathrm{AB}}}{|\overline{\mathrm{AB}}|}\) is the unit vector along \(\overleftrightarrow{\mathrm{AB}}\) in the direction from A to B.

If A = (x₁, y₁, z₁), B = (x₂, y₂, z₂) then the unit vector along \(\overleftrightarrow{\mathrm{AB}}\) in the direction from A to B = \(\frac{\left(x_2-x_1, y_2-y_1, z_2-z_1\right)}{\sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2\right]}}\)

If \(\overline{\mathrm{OA}}=\bar{a}, \overline{\mathrm{OB}}=\bar{b}\) are two non-collinear vectors, \(\overleftrightarrow{\mathrm{OA}}, \overleftrightarrow{\mathrm{OB}}\) determine a unique plane denoted by \(\overleftrightarrow{\mathrm{AOB}}\) and we say that it is the plane containing \(\bar{a}, \bar{b}\).

Coordinates Coplanar, Non-Coplanar Vectors

Let \(\bar{a}, \bar{b}\) be two non-collinear vectors and \(\bar{c}\) be a vector. Let O be the origin and A, B, C be three points such that \(\overline{\mathrm{OA}}=\bar{a}, \overline{\mathrm{OB}}=\bar{b}, \overline{\mathrm{OC}}=\bar{c}\). Since \(\overline{\mathrm{OA}}, \overline{\mathrm{OB}}\) are non-collinear, they determine the plane \(\overleftrightarrow{\mathrm{AOB}}\).

If C ∈ \(\overleftrightarrow{\mathrm{AOB}}\), then \(\overline{\mathrm{OA}}, \overline{\mathrm{OB}}, \overline{\mathrm{OC}}\) are said to be coplanar and if \(\mathrm{C} \notin \overleftrightarrow{\mathrm{AOB}}\), then \(\overline{\mathrm{OA}}, \overline{\mathrm{OB}}, \overline{\mathrm{OC}}\) are said to be non-coplanar. \(\bar{i}, \bar{j}, \bar{k}\) are non-coplanar.

Coordinates Angle Between Vectors

If \(\bar{a}=\overline{\mathrm{OA}}, \bar{b}=\overline{\mathrm{OB}}\), then the angle between the vectors \(\bar{a}, \bar{b}\) [written as (\(\bar{a}, \bar{b}\))] is (\(\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}\)) such that \(0^{\circ} \leq(\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}) \leq 180^{\circ}\).

We write \(0^{\circ} \leq(\bar{a}, \bar{b}) \leq 180^{\circ}\). We have \((\bar{a},-\bar{b})=(-\bar{a}, \bar{b})=(\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{BO}})=(\overrightarrow{\mathrm{AO}}, \overrightarrow{\mathrm{OB}})\)

If \((\bar{a}, \bar{b})=90^{\circ}\), then we write \(\bar{a} \perp \bar{b} \text { or } \overline{\mathrm{OA}} \perp b \text { or } \bar{a} \perp \overline{\mathrm{OB}} \text { or } \overline{\mathrm{OA}} \perp \overline{\mathrm{OB}}\).

We take that the null vector is perpendicular to every vector.

If \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar and \(\bar{r}\) is any vector, then there exist unique real numbers x, y, z such that \(\bar{r}=x \bar{a}+y \bar{b}+z \bar{c}\). \(\bar{r}\) is said to be a linear combination of \(\bar{a}, \bar{b}, \bar{c}\).

Since \(\bar{i}, \bar{j}, \bar{k}\) are non-coplanar \(\bar{r}=x \bar{a}+y \bar{b}+z \bar{c}\). \(\bar{r}\), for unique scalars x, y, z.

⇒ \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar.

(1) \(x_1 \bar{a}+y_1 \bar{b}+z_1 \bar{c}=x_2 \bar{a}+y_2 \bar{b}+z_2 \bar{c} \Rightarrow x_1=x_2, y_1=y_2, z_1=z_2\)

(2) \(x_1 \bar{a}+y_1 \bar{b}+z_1 \bar{c}=0 \Rightarrow x_1=0, y_1=0, z_1=0\).

Solved Problems On Interpretation Of Equations In Plane Coordinates

Definition. A, B are two points. If λ₁, λ₂,(λ₁ + λ₂ ≠ 0) are two real numbers such that P ∈ AB and \(\lambda_2 \overline{\mathrm{AP}}=\lambda_1 \overline{\mathrm{PB}}\), we say that P divides AB in the ratio λ₁ : λ₂.

λ₁ : λ₂ and λ₁, λ₂ < 0, P-A-B or A-B-P i.e., P is said to divide the line segment AB externally in the ratio λ₁:λ₂.

If \(\mathrm{A}=\bar{a}, \mathrm{~B}=\bar{b} \text { and }(\mathrm{P} ; \mathrm{A}, \mathrm{B})=\lambda_1: \lambda_2 \text { then } \overline{\mathrm{OP}}=\frac{\lambda_1 \bar{a}+\lambda_2 \bar{b}}{\lambda_1+\lambda_2}\left(\lambda_1+\lambda_2 \neq 0\right)\)

⇒ \(\overline{\mathrm{A}}, \overline{\mathrm{B}}, \overline{\mathrm{C}}\) are three points whose position vectors are \(\bar{a}, \bar{b}, \bar{c}\) respectively. \(\mathrm{A}(\bar{a}), \mathrm{B}(\bar{b}), \mathrm{C}(\bar{c})\) are collinear <=> \(\overline{l a}+m \bar{b}+n \bar{c}=0, l+m+n=0,(l, m, n) \neq(0,0,0)\)

Coordinates Dot Product Or Direct Product Or Scalar Product Or Inner Product.

Definition. If \(\vec{a}, \vec{b}\) are two non-zero vectors, then \(\bar{a} \cdot \bar{b}=|\bar{a}||\bar{b}| \cos (\bar{a}, \bar{b})\) and if one of \(\bar{a} \cdot \bar{b}\) is zero then \(\bar{a} \cdot \bar{b}=0\).

Now \(\bar{i} \cdot \bar{i}=1, \bar{j} \cdot \bar{j}=1, \bar{k} \cdot \bar{k}=1, \bar{i} \cdot \bar{j}=0, \bar{i} \cdot \bar{k}=0, \bar{j} \cdot \bar{k}=0\).

If \(\bar{a}=\left(x_1, y_1, z_1\right), \bar{b}=\left(x_2, y_2, z_2\right) \text {, then } \bar{a} \cdot \bar{b}=\left(x_1, y_1, z_1\right) \cdot\left(x_2, y_2, z_2\right)\)

= \(\left(x_1 \bar{i}, y_1 \bar{j}, z_1 \bar{k}\right) \cdot\left(x_2 \bar{i}, y_2 \bar{j}, z_2 \bar{k}\right)=x_1 x_2+y_1 y_2+z_1 z_2\)

Also we have (x₁, y₁, z₁).(x₂, y₂, z₂) = (x₂, y₂, z₂).(x₁, y₁, z₁)

Also if \(|\bar{a}|=\bar{a} \text {, then } \bar{a}^2=\bar{a}^2=\bar{a} \cdot \bar{a}=x_1^2+y_1^2+z_1^2\)

If \(\bar{a}\) is a unit vector, then \(\bar{a}^2=|\bar{a}|^2=1\).

If \(\mathrm{P}=\bar{a}=\left(a_1, b_1, c_1\right), \mathrm{Q}=\bar{b}=\left(a_2, b_2, c_2\right), \mathrm{P} \neq \mathrm{Q} \neq \mathrm{O} \text { and }(\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}})=(\bar{a}, \bar{b})=\theta\)

then \(\cos \theta=\frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|}=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)} \cdot \sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}\)

⇒ \(\bar{a} \cdot \bar{b}\) are parallel vectors <=> \(\bar{a}=\lambda \bar{b}\)

⇒ \(\left(a_1, b_1, c_1\right)=\lambda\left(a_2, b_2, c_2\right) \Leftrightarrow a_1=\lambda a_2, b_1=\lambda b_2, c_1=\lambda c_2\)

⇒ \(a_1: b_1: c_1=a_2: b_2: c_2 \text { or } a_1: a_2=b_1: b_2=c_1: c_2\)

⇒ \(\bar{a} \cdot \bar{b}\) are perpendicular vectors <=> \(\bar{a} \cdot \bar{b}=0 \Leftrightarrow a_1 a_2+b_1 b_2+c_1 c_2=0\)

Projection of \(\bar{b} \text { on } \bar{a}(\neq 0) \text { is } \bar{b} \cdot \frac{\bar{a}}{|\bar{a}|}=\frac{\bar{b} \cdot \bar{a}}{|\bar{a}|}=\bar{b} \cdot \bar{e}\) where \(\bar{e}\) is the unit vector in the direction of \(\bar{a}\).

⇒ \(\bar{a} \cdot(\bar{b}+\bar{c})=\bar{a} \cdot \bar{b}+\bar{a} \cdot \bar{c}, \bar{a} \cdot(\bar{b}-\bar{c})=\bar{a} \cdot \bar{b}-\bar{a} \cdot \bar{c}\)

⇒ \((\bar{a}-\bar{b})^2=(\bar{a}-\bar{b}) \cdot(\bar{a}-\bar{b})=\bar{a}^2-2 \bar{a} \cdot \bar{b}+\bar{b}^2\)

Coordinates Cross Product Skew Product Or Vector Product

If \(\bar{a}, \bar{b}\) are two non-zero or non-parallel vectors, then \((\bar{a} \times \bar{b})=|\bar{a}||\bar{b}| \sin (\bar{a}, \bar{b}) \bar{n}\) where \(\bar{n}\) is a unit vector perpendicular to the plane containing \(\bar{a}, \bar{b}\) so that \(\bar{a}, \bar{b}, \bar{n}\) form a right handed system and if at least one of \(\bar{a}, \bar{b}\) is a null vector or \(\bar{a} \| \bar{b}\), then \((\bar{a} \times \bar{b})=0\).

We find that \(\bar{a} \times \bar{b}=-\bar{b} \times \bar{a} \text { and }|\bar{a} \times \bar{b}|=|\bar{a}||\bar{b}||\sin (\bar{a}, \bar{b})|\)

We find that \(\bar{a} \times \bar{b}=-\bar{b} \times \bar{a} \text { and }|\bar{a} \times \bar{b}|=|\bar{a}||\bar{b}||\sin (\bar{a}, \bar{b})|\)

Now \(\bar{i} \times \bar{i}=0, \bar{j} \times \bar{j}=0, \bar{k} \times \bar{k}=0, \bar{i} \times \bar{j}=\bar{k}, \bar{i} \times \bar{k}=-\bar{j}, \bar{j} \times \bar{k}=\bar{i}, \bar{k} \times \bar{j}=-\bar{i}\)

⇒ \(\bar{j} \times \bar{i}=\bar{k}, \bar{k} \times \bar{i} \times \bar{j} .\)

If \(\bar{a}=\left(x_1, y_1, z_1\right), \bar{b}=\left(x_2, y_2, z_2\right)\), then

= \(x_1 y_2 \bar{k}-x_1 z_2 \bar{j}-x_2 y_1 \bar{k}+y_1 z_2 \bar{i}+z_1 x_2 \bar{j}-z_1 y_2 \bar{i}\)

= \(\left(y_1 z_2-y_2 z_1\right) \bar{i}-\left(x_1 z_2-x_2 z_1\right) \bar{j}+\left(x_1 y_2-x_2 y_1\right) \bar{k}\)

= \(\left(y_1 z_2-y_2 z_1, x_2 z_1-x_1 z_2, x_1 y_2-x_2 y_1\right)=\left|\begin{array}{ccc}
\bar{i} & \bar{j} & \bar{k} \\
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2
\end{array}\right|\)

i.e. \(\left(x_1, y_1, z_1\right) \times\left(x_2, y_2, z_2\right)=\left(y_1 z_2-y_2 z_1, x_2 z_1-x_1 z_2, x_1 y_2-x_2 y_1\right)\)

⇒ \(\left(x_2, y_2, z_2\right) \times\left(x_1, y_1, z_1\right)=-\left(y_1 z_2-y_2 z_1, x_2 z_1-x_1 z_2, x_1 y_2-x_2 y_1\right)\)

and \(\left|\left(x_1, y_1, z_1\right) \times\left(x_2, y_2, z_2\right)\right|=\sqrt{\left[\sum\left(y_1 z_2-y_2 z_1\right)^2\right]}\)

If \(\mathrm{P}=\bar{a}=\left(a_1, b_1, c_1\right), \mathrm{Q}=\bar{b}=\left(a_2, b_2, c_2\right)=(\mathrm{P} \neq \mathrm{Q} \neq \mathrm{O})\)

and \((\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}})=(\vec{a}, \bar{b})=\theta\), then

⇒ \(\sin \theta=\frac{|\bar{a} \times \bar{b}|}{|\bar{a}||\bar{b}|}=\frac{\left|\left(b_1 c_2-b_2 c_1, c_1 a_2-c_2 a_1, a_1 b_2-a_2 b_1\right)\right|}{\sqrt{\left(a_1^2+b_1^2+c_1^2\right)} \cdot \sqrt{\left(a_2^2+b_2^2+c_2^2\right)}}\)

If  \(\overline{\mathrm{ABC}}\) is a triangle, then the area of \(\Delta \overline{\mathrm{ABC}}=\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|\) square units.

Also \(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}\) is a vector perpendicular to the plane of \(\triangle \overline{\mathrm{ABC}}\).

Area of \(\triangle \overline{\mathrm{ABC}}\) = 0 ⇔ A, B, C are collinear.

A, B, C, and D are coplanar points. If ABCD is a parallelogram then the area of the parallelogram = \(|\overline{\mathrm{AB}} \times \overline{\mathrm{AD}}| \text { or } \frac{1}{2}|\overline{\mathrm{AC}} \times \overline{\mathrm{BD}}|\) square units.

If ABCD is a quadrilateral, then the area of the quadrilateral = \(\frac{1}{2}|\overline{\mathrm{AC}} \times \overline{\mathrm{BD}}|\) square units.

⇒ \(\bar{a}=\left(a_1, a_2, a_3\right), \bar{b}=\left(b_1, b_2, b_3\right), \bar{c}=\left(c_1, c_2, c_3\right)\)

⇒ \(\bar{a} \times \bar{b}=\left(a_2 b_3-a_3 b_2, a_3 b_1-a_1 b_3, a_1 b_2-a_2 b_1\right)\)

∴ \([\bar{a} \bar{b} \bar{c}]=(\bar{a} \times \bar{b}) \cdot \bar{c}\)

= \(\left(a_2 b_3-a_3 b_2, a_3 b_1-a_1 b_3, a_1 b_2-a_2 b_1\right) \cdot\left(c_1, c_2, c_3\right)\)

= \(\left(a_2 b_3 c_1-a_3 b_1 c_1+a_3 b_1 c_2-a_1 b_3 c_2+a_1 b_2 c_3-a_2 b_1 c_3\right)=\left|\begin{array}{ccc}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|\)

i.e. \([\bar{a} \bar{b} \bar{c}]=\left[\left(a_1, a_2, a_3\right),\left(b_1, b_2, b_3\right),\left(c_1, c_2, c_3\right)\right]=\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|\)

⇒ \(\bar{a}, \bar{b}, \bar{c}\) are three non-coplanar vectors. If V is the volume of the parallelopiped with adjacent sides \(\bar{a}, \bar{b}, \bar{c}\) then  \( \bar{V}=|[\bar{a}, \bar{b}, \bar{c}]|\) cubic units.

If V is the volume of the tetrahedron with adjacent sides \(\bar{a}, \bar{b}, \bar{c}\) then \(\overline{\mathrm{V}}=\frac{1}{6}|[\bar{a}, \bar{b}, \bar{c}]|\) cubic units. Also if any two of \(\bar{a}, \bar{b}, \bar{c}\) are parallel, \(\lceil\bar{a}, \bar{b}, \bar{c}\rceil=0\)

One of \(\bar{a}, \bar{b}, \bar{c}\) is \(0 \Rightarrow[\bar{a}, \bar{b}, \bar{c}]=0\) . Also if any two \(\bar{a}, \bar{b}, \bar{c}\) are parallel, \([\bar{a}, \bar{b}, \bar{c}]=0\).

⇒ \(\bar{a}, \bar{b}, \bar{c}\) are three non-zero, non-parallel vectors. \(\bar{a}, \bar{b}, \bar{c}\) are parallel, <=> \([\bar{a}, \bar{b}, \bar{c}]=0\).

If \(\bar{a}, \bar{b}, \bar{c}\) are three non-coplanar vectors, then \([\bar{a}, \bar{b}, \bar{c}]^2=\left|\begin{array}{lll}
\bar{a} \cdot \bar{a} & \bar{a} \cdot \bar{b} & \bar{a} \cdot \bar{c} \\
\bar{b} \cdot \bar{a} & \bar{b} \cdot \bar{b} & \bar{b} \cdot \bar{c} \\
\bar{c} \cdot \bar{a} & \bar{c} \cdot \bar{b} & \bar{c} \cdot \bar{c}
\end{array}\right|\)

A, B are two distinct points. Distance of P from  \(\overleftrightarrow{\mathrm{AB}}=\frac{|\overline{\mathrm{AP}} \times \overline{\mathrm{AB}}|}{|\overline{\mathrm{AB}}|}\)

A, B, and C are distinct points.

⇒ \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}\) are parallel <=> A, B, C are collinear.

A, B, C, and D are distinct points.

⇒ \(\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}\) are coplanar <=> A, B, C, D are coplanar

Coordinates Notation   

⇒ \(‘ \bar{m} \perp \overline{\mathrm{L}} \text { ‘ means : }\) There exist two points A, B such that \(\overline{\mathrm{AB}}=\bar{m} \text { and } \mathrm{AB} \perp \mathrm{L} \text {. }\)

⇒ \(‘ \bar{m} \| \overline{\mathrm{L}} \text { ‘ means: }\) There exist two points A, B such that \(\overline{\mathrm{AB}}=\bar{m} \text { and } \mathrm{AB} \| \mathrm{L}\)

⇒ \({ }^{\prime} \bar{m} \subset \pi^{\prime} \text { means: }\) There exist two points A, B such that \(\overline{\mathrm{AB}}=\bar{m} \text { and } \overleftrightarrow{\mathrm{AB}} \subset \pi\)

⇒ \({ }^{\prime} \bar{m} \subset \pi^{\prime} \text { means: }\) \(\overline{\mathrm{m}}\) is a point π.

Coordinates Solved Problems

Example.1. Show that the points (3, -2, 4),(1, 1, 1), (-1, 4, -2) are collinear. Hence find (C; A, B).

Solution.

Given

(3, -2, 4),(1, 1, 1), (-1, 4, -2)

Let A = (3,-2,4), B = (1,1,1) and C = (-1,4,-2)

∴ \(\mathrm{AB}=|\overline{\mathrm{AB}}|=|(1-3,1+2,1-4)|=|(-2,3,-3)|=\sqrt{(4+9+9)}=\sqrt{(22)}\),

BC \(=|(-2,3,-3)|=\sqrt{(4+9+9)}=\sqrt{(22)}\)

AC\(=|(-4,6,-6)|=\sqrt{(16+36+36)}=2 \sqrt{(22)} .\)

∴ AB + BC = AC.

∴ A, B, and C are collinear, and (C, A, B) = (-1-3):(1+1) = -2:1

⇒ \(\mathrm{OR}: \overline{\mathrm{AC}}=(-4,6,-6), \overline{\mathrm{CB}}=(2,-3,3)\)

Since \(\overline{\mathrm{AC}}=-2(2,-3,3)=-2 \overline{\mathrm{CB}}\) A, B, C are collinear.

Let (C, A, B) = λ1 : λ2

∴ \(\lambda_2 \overline{\mathrm{AC}}=\lambda_1 \overline{\mathrm{CB}}\)

⇒ \(\lambda_2(-4,6,-6)=\lambda_1(2,-3,3) \Rightarrow 2 \lambda_1=-4 \lambda_2\)

⇒ \(\lambda_1: \lambda_2=-2: 1\) ⇒ (C, A, B) = -2:1

Example.2. Show that the points (-1, -2, -1), (2, 3, 2), (4, 7, 6), and (1, 2, 3) form a parallelogram.

Solution.

Given

(-1, -2, -1), (2, 3, 2), (4, 7, 6), and (1, 2, 3)

Let A = (-1, -2, -1), B = (2, 3, 2), C = (4, 7, 6) and D = (1, 2, 3)

∴ \(\mathrm{AB}=\sqrt{\left[(2+1)^2+(3+2)^2+(2+1)^2\right]}=\sqrt{(43)}, \mathrm{BC}=6, \mathrm{CD}=\sqrt{(43)}, \mathrm{AD}=6\)

Also \(\mathrm{AC}=\sqrt{(155)} \text { and } \mathrm{BD}=\sqrt{3}\)

∴ AB = CD, BC = AD and AC ≠ BD.

Example.3. Find the center and radius of the sphere determined by the points (1, -5, -3), (0, -6, -1), (-2, -2, 3), (1, -2, 0).

Solution.

Given

(1, -5, -3), (0, -6, -1), (-2, -2, 3), (1, -2, 0)

The sphere is the set of points P = (x, y, z) where A = (a, b, c) and PA = r (a non-negative number) A is called the center and r is called the radius.

Let \(P_1=(1,-5,3), P_2=(0,-6,-1), P_3=(-2,-2,3) \text { and } P_4=(1,-2,0) \text {. }\)

∴ \(P_1 A=P_2 A=P_3 A=P_4 A \Rightarrow P_1 A^2=P_2 A^2=P_3 A^2=P_4 A^2\)

⇒ \((a-1)^2+(b+5)^2+(c-3)^2=(a-0)^2+(b+6)^2+(c+1)^2\)   ………(1)

= \((a+2)^2+(b+2)^2+(c-3)^2=(a-1)^2+(b+2)^2+(c-0)^2\)   ……..(2)

From (1)  and (2): -2a – 2b – 8c = 2 ⇒ a + b + 4c = -1  …….(3)

-6a + 6b = -18 ⇒ a – b = 3

6b – 6c = -30 ⇒ b – c = -5

Solving 1 and 2 a = -1, b = -4, c = 1.

∴ Centre A = (-1, -4, 1).

Radius \(\mathrm{P}_1 \mathrm{~A}=\sqrt{\left[(a-1)^2+(b+5)^5+(c-3)^2\right]}=\sqrt{(4+1+4)}=3\)

Theorem. 2 If \(\overline{\mathrm{A}}=\left(x_1, y_1, z_1\right), \overline{\mathrm{B}}=\left(x_2, y_2, z_2\right)\) and P is a point the line segment \(\mathrm{AB}\) in the ratio λ₁ : λ₂ (λ₁ + λ₂ ≠ 0), then \(P=\left(\frac{\lambda_2 \bar{x}_1+\lambda_1 \bar{x}_2}{\lambda_1+\lambda_2} \cdot \frac{\lambda_2 \bar{y}_1+\lambda_1 \bar{y}_1}{\lambda_1+\lambda_2} \cdot \frac{\lambda_2 \bar{z}_1+\lambda_1 \bar{z}_1}{\lambda_1+\lambda_2}\right)\)

Proof: Let P = (x, y, z).

A, P, B = \(\lambda_1: \lambda_2\left(\lambda_1+\lambda_2 \neq 0\right)\)

⇔ \(\lambda_2 \overline{\mathrm{AP}}: \lambda_1 \overline{\mathrm{PB}}\)

⇔ \(\lambda_2\left(x-x_1, y-y_1, z-z_1\right)=\lambda_1\left(x_2-x, y_2-y, z_2-z\right)\)

⇔ \(\lambda_2\left(x-x_1\right)=\lambda_1\left(x_2-x\right)\), etc.

⇔ \(\left(\lambda_1+\lambda_2\right) x=\lambda_2 x_1+\lambda_1 x_2\), etc.

⇔ \(x=\frac{\lambda_2 x_1+\lambda_1 x_2}{\lambda_1+\lambda_2}, y=\frac{\lambda_2 y_1+\lambda_1 y_2}{\lambda_1+\lambda_2}, z=\frac{\lambda_2 z_1+\lambda_1 z_2}{\lambda_1+l_2}\)

⇔ \(\mathrm{P}=\left(\frac{\lambda_2 x_1+\lambda_1 x_2}{\lambda_1+\lambda_2}, \frac{\lambda_2 y_1+\lambda_1 y_2}{\lambda_1+\lambda_2}, \frac{\lambda_2 z_1+\lambda_1 z_2}{\lambda_1+I_2}\right),\left(\lambda_1+\lambda_2 \neq 0\right)\)

Note. 1. If λ₁ = λ₂, P wil be mid-point of AB and \(\mathrm{P}=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)\)

2. \(\mathrm{A}=\left(x_1, y_1, z_1\right), \mathrm{B}=\left(x_2, y_2, z_2\right) \text {, }\)

P \(=\left(\frac{\lambda_2 x_1+\lambda_1 x_2}{\lambda_1+\lambda_2}, \frac{\lambda_2 y_1+\lambda_1 y_2}{\lambda_1+\lambda_2}, \frac{\lambda_2 z_1+\lambda_1 z_2}{\lambda_1+\lambda_2}\right)\)

are three points and λ₁, λ₂ ∈ R such that λ₁ + λ₂ ≠ 0.

⇒ A, P, B are collinear.

Theorem 3. If (x_r, y_r, z_r), r = 1, 2, 3 are the vertices of a triangle, then its medians are concurrent and the point of concurrence trisects any median of the triangle.

Proof.

Given

If (x_r, y_r, z_r), r = 1, 2, 3 are the vertices of a triangle

Let ABC be the triangle where A = (x₁, y₁, z₁), B = (x₂, y₂, z₂), C = (x₃, y₃, z₃)

Let D, E, and F be the mid-points of the sides. AD, BE, and CF are the medians of ΔABC.

Now D = \(\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}, \frac{z_2+z_3}{2}\right)\)

Let (G, A, D) = 2: 1.

∴ \(\mathrm{G}=\frac{2\left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}, \frac{z_2+z_3}{2}\right)+1\left(x_1, y_1\right)}{2+1}\)

= \(\frac{\left(x_1+x_2+x_3, y_1+y_2+y_3, z_1+z_2+z_3\right)}{3}\)

= \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)\)

Similarly, the point dividing BE in the ratio 2:1 is

⇒ \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)\) and the point dividing CF in the ratio 2: 1 is \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)\)

∴ G is the point common to AD, BE, and CF.

∴ Medians in a triangle are concurrent and the point of concurrence trisects each median. This point G is called the centroid.

Coordinates Tetrahedron

Let A, B, C, D be four points such that \(\overleftrightarrow{\mathrm{ABC}}, \overleftrightarrow{\mathrm{ABD}}, \overleftrightarrow{\mathrm{ADC}}, \overleftrightarrow{\mathrm{BCD}}\) are four intersecting planes.

Then points A, B, C, and D are said to form a tetrahedron. A, B, C, and D are called vertices and the line segments AB, AD, AC, BC, BD, CD are called the edge, AB,  CD; AD, BC; AC, BD are called three pairs of opposite edges.

Observe that for the tetrahedron:

(1) Each of the points A, B, C, and D is non-coplanar with the remaining three.

(2) Opposite edges from non-coplanar lines i.e., \(\overleftrightarrow{\mathrm{AB}}, \overleftrightarrow{\mathrm{CD}}\) are non-coplanar, \(\overleftrightarrow{\mathrm{AD}}, \overleftrightarrow{\mathrm{BC}}\) are non-coplanar; \(\overleftrightarrow{\mathrm{AC}}, \overleftrightarrow{\mathrm{BD}}\) are non-coplanar.

(3) Four bounding planes \(\overleftrightarrow{\mathrm{ABD}}, \overleftrightarrow{\mathrm{ADC}}, \overleftrightarrow{\mathrm{ABC}}, \overleftrightarrow{\mathrm{BCD}}\) are triangular faces.

The point of concurrence of the line segments joining the vertices to their respective centroids of opposite triangular faces is called the centroid of the tetrahedron.

If all the edges are of equal length, then it is called a regular tetrahedron.

Theorem.4. If A = (x₁, y₁, z₁), B = (x₂, y₂, z₂), C = (x₃, y₃, z₃), D = (x₄, y₄, z₄) are the vertices of the tetrahedron ABCD, then the line segments joining the vertices to their respective centroids of opposite faces are concurrent and the point of concurrence divides each line segment in the ratio 3: 1.

Proof. Let S, P, Q, and R be the centroids of

⇒ \(\triangle \mathrm{BCD}, \triangle \mathrm{ACD}, \triangle \mathrm{ABD}, \triangle \mathrm{ABC}\) respectively.

∴ \(\S=\left(\frac{x_2+x_3+x_4}{3}, \frac{y_2+y_3+y_4}{3}, \frac{z_2+z_3+z_4}{3}\right)\)

Let G divide the line segments AS in the ration 3:1. Since A = (x₁, y₁, z₁),

G = \(\left(\frac{3\left(\frac{x_2+x_3+x_4}{3}\right)+1 \cdot x_1}{3+1}, \frac{3\left(\frac{y_2+y_3+y_4}{3}\right)+1 \cdot y_1}{3+1}, \frac{3\left(\frac{z_2+z_3+z_4}{3}\right)+1 \cdot z_1}{3+1}\right)\)

i.e., G = \(\left(\frac{\left(x_1+x_2+x_3+x_4\right)}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)\)

Similarly, the points that divide the other line segments BP, CQ, and DR in the ratio 3: 1 can be shown to be G.

The line segments AS, BP, CQ, and DR are concurrent at G and each is divided in the ratio 3: 1 at G. G is called the centroid of the tetrahedron.

Example.1 Show that the following points are collinear, find A = (5, 4, 2), B = (8, -2, -7), C = (6, 2, -1) If A. B. C are collinear, find (C ; A, B)

Solution.

Given

A = (5, 4, 2), B = (8, -2, -7), C = (6, 2, -1)

Let P divide AB in the ratio 1 : λ (1 + λ ≠ 0).

∴ P = \(\left(\frac{5 \lambda+8}{1+\lambda}, \frac{-2+4 \lambda}{1+\lambda}, \frac{-7+2 \lambda}{1+\lambda}\right)\)

If possible, let P = C.

Then \(\left.\begin{array}{c}
\frac{5 \lambda+8}{1+\lambda}=6, \\
\frac{-2+4 \lambda}{1+\lambda}=2, \\
\frac{-7+2 \lambda}{1+\lambda}=-1
\end{array}\right\}\)

∴ λ = 2.

Since λ = 2 satisfies all three equations, A, B, and C are collinear.

∴ (C; A, B) = 1: 2.

Coordinates Direction Cosines Of A Line

\(\overrightarrow{\mathrm{PQ}}\) is a ray making angles α,β, γ respectively with \(\overrightarrow{\mathrm{OX}}\), \(\overrightarrow{\mathrm{OY}}, \overrightarrow{\mathrm{OZ}}\).

Then the ordered triad \((\cos \alpha, \cos \beta, \cos \gamma)\) is called direction cosines triad of \(\overrightarrow{\mathrm{PQ}}\) i.e., \(\cos \alpha, \cos \beta, \cos \gamma\) in that order are called the direction cosines (d. cs.) of \(\overrightarrow{\mathrm{PQ}}\)

The direction cosine triad is generally denoted by (l, m, n). Thus cosα = 1, cosβ = m, cosγ = n, and the d.cs. of are l, m, n.

Since the ray \(\overrightarrow{\mathrm{QP}}\) makes angles 180° – α, 180° – β, 180° – γ respectively with \(\overrightarrow{\mathrm{OX}}, \overrightarrow{\mathrm{OY}}, \overrightarrow{\mathrm{OZ}}\) direction cosine triad of \(\overrightarrow{\mathrm{QP}}\) is [(cos(180° – α), cos(180° – β), cos(180° – γ)]

i.e. (-cosα, -cosβ, -cosγ) i.e. (-l, -m, -n)

If L is a line parallel to then the two ordered triads (l, m, n) in that order and -l, -m, -n in that order are defined as the d.c.s of L. Sometimes d. cs. l, m, n are written as (l, m, n).

Theorem.5. If l, m, n are d.cs of a line, then l² + m² + n² = 1.

Proof:

Let L be the line with d.cs. l, m, n

∴ (cos α, cos β, cos γ) = (l, m, n) or (-l, -m, -n)

If P(x, y, z) (≠G) is a point such that \(\overleftrightarrow{\mathrm{OP}} \| \mathrm{L}\) and OP = 1, then:

From Trigonometry

In \(\overleftrightarrow{\mathrm{POX}}\) plane, \(\cos \alpha=\frac{x}{1}=x\);

In \( \overleftrightarrow{\mathrm{POY}}\) plane, \(\cos \beta=\frac{y}{1}=y\);

In \(\overleftrightarrow{\mathrm{POZ}}\) plane, \(\cos \gamma=\frac{z}{1}=z\);

OR

a = \((\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OX}})=(\overline{\mathrm{OP}}, \bar{i}) \Rightarrow \cos \alpha=\overline{\mathrm{OP}}, \bar{i}=(x, y, z) \cdot(1,0,0)=x \text {, etc. }\)

i.e. (x, y, z) = (l, m, n) or (-l, -m, -n)

But \(\mathrm{OP}^2=1 \Rightarrow x^2+y^2+z^2=1 \Rightarrow l^2+m^2+n^2=1 \text { etc. }\)

Note. (l, m, n) and (-l, -m, -n) are the only unit points on \(\stackrel{\leftrightarrow}{\mathrm{OP}}\) and l, m, n; -l, -m, -n are the d.cs. of L.

Coordinates Direction Ratios Or Direction Numbers Of A Line

L is a line and p(x, y, z) is a point such that \(\overleftrightarrow{\mathrm{OP}} \| \mathrm{L}\). Then the coordinates of any point on \(\overleftrightarrow{\mathrm{OP}}\), other than the origin, are called the direction numbers of L. But any point, other than the origin, on \(\overleftrightarrow{\mathrm{OP}}\) is (λx, λy, λz)(λ≠0). So in that order are called the direction numbers of the line L.

Clearly, the direction numbers for a line L are infinitely many and they are proportional. Further, the direction numbers cannot be 0, 0, 0.

The direction numbers are sometimes called as direction ratios(d. rs.)

The d.cs. (l, m, n) or (-l, -m, -n) of L are also d.rs. of L since each of (l, m, n) or (-l, -m, -n) is also a point on \(\overleftrightarrow{\mathrm{OP}}\).

If P(x, y, z) then unit points on \(\overleftrightarrow{\mathrm{OP}}\) are

⇒ \(\left(\frac{x}{\sqrt{\left(x^2+y^2+z^2\right)}}, \frac{y}{\sqrt{\left(x^2+y^2+z^2\right)}}, \frac{z}{\sqrt{\left(x^2+y^2+z^2\right)}}\right)\),

⇒  \(\left(\frac{-x}{\sqrt{\left(x^2+y^2+z^2\right)}}, \frac{-y}{\sqrt{\left(x^2+y^2+z^2\right)}}, \frac{-z}{\sqrt{\left(x^2+y^2+z^2\right)}}\right)\)

Hence if d.rs. of L are x, y, z then d.cs. of L are

⇒  \(\frac{x}{\sqrt{\left(x^2+y^2+z^2\right)}}, \frac{y}{\sqrt{\left(x^2+y^2+z^2\right)}}, \frac{z}{\sqrt{\left(x^2+y^2+z^2\right)}}\)

⇒  \(\frac{-x}{\sqrt{\left(x^2+y^2+z^2\right)}}, \frac{-y}{\sqrt{\left(x^2+y^2+z^2\right)}}, \frac{-z}{\sqrt{\left(x^2+y^2+z^2\right)}}\)

Note.1. D.rs. of the coordinate axes are respectively x, 0, 0; 0, y, 0; 0, 0, z (x≠0, y≠0, z≠0)

2. If x, y, z are d.rs. of a line L, there exists a point p(x, y, z) such that \(\overleftrightarrow{\mathrm{OP}} \| \mathrm{L}\) and \(x \bar{i}+y \bar{j}+z \bar{k}\) is a vector along \(\overleftrightarrow{\mathrm{OP}}\) i.e. a vector along L.

3. If l, m, n are the d.cs. of a line L then \(l \bar{i}+m \bar{j}+n \bar{k}\) is a unit vector L.

4. l, m, n are d.cs. of a line L. If any two of the d.cs. and the sign of the third is known, then the d.cs. of L can be found.

For example, \(\frac{1}{2}, \frac{1}{2}, n\) are d.cs. of L and n < 0 ⇒ \(n=-\sqrt{\left(1-\frac{1}{4}-\frac{1}{4}\right)}=\frac{1}{\sqrt{2}}\)

⇒ d.cs. of L are \(\frac{1}{2}, \frac{1}{2},-\frac{1}{\sqrt{2}}\)

example. The d.cs. of the line with d.rs. (3, 2, 6) are

± \(\frac{3}{\sqrt{(9+4+36)}}, \pm \frac{2}{(9+4+36)}, \pm \frac{6}{\sqrt{(9+4+36)}}\)

i.e. \(\frac{3}{7}, \frac{2}{7}, \frac{6}{7} ;-\frac{3}{7},-\frac{2}{7},-\frac{6}{7}\)

Theorem.6. If P=(x₁, y₁, z₁), Q = (x₂, y₂, z₂) then x₂ – x₁, y₂ – y₁, z₂ – z₁ are d.rs. of \(\overline{\mathrm{PQ}}\).

Proof. Let OPQR be a parallelogram

Then \(\overline{\mathrm{OR}}=\overline{\mathrm{PQ}}=\left(x_2-x_1, y_2-y_1, z_2-z_1\right)\)

⇒ \(\mathrm{R}=\left(x_2-x_1, y_2-y_1, z_2-z_1\right)\)

∴ d.rs. of \(\overleftrightarrow{\mathrm{OR}} \text { are } x_2-x_1, y_2-y_1, z_2-z_1\)

∴ d.rs. of \(\overleftrightarrow{\mathrm{PQ}} \text { are } x_2-x_1, y_2-y_1, z_2-z_1\)

Coordinates Projection Of A Line Segment On Another Line

The projection of a line segment CD on a line \(\overleftrightarrow{\mathrm{AB}}\)  is MN where M, N are the feet of the perpendiculars on \(\overleftrightarrow{\mathrm{AB}}\).

If \(\overleftrightarrow{\mathrm{CD}}\) make an angle θ with \(\overleftrightarrow{\mathrm{AB}}\), then the projection of CD on \(\overleftrightarrow{\mathrm{AB}}\) is MN = CDcosθ.

Let \(\overleftrightarrow{\mathrm{AB}}\) ne a line. On \(\overleftrightarrow{\mathrm{AB}}\) let the projection of C, and D be M and, N respectively.

Then the projection of CD on the line \(\overleftrightarrow{\mathrm{AB}}\) is \(\overline{\mathrm{CD}}\). \(\frac{\overline{\mathrm{MN}}}{|\overline{\mathrm{MN}}|}\) in the direction \(\overline{\mathrm{MN}}\).

Theorem.7. If \(\overleftrightarrow{\mathrm{AB}}\) is a ray with d.cs. l, m, n, and P = (x₁, y₁, z₁), Q = (x₂, y₂, z₂) are two points, then the projection of PQ on \(\overleftrightarrow{\mathrm{AB}}\) in the direction \(\overleftrightarrow{\mathrm{AB}}\) is (x₂ – x₁)l + (y₂ – y₁)m + (z₂ – z₁)n.

Proof.

PQ = \(\left(x_2-x_1, y_2-y_1, z_2-z_1\right)\)

Unit vector along \(\overrightarrow{\mathrm{AB}}\) = e = (l, m, n)

∴ Projection PQ on \(\overleftrightarrow{\mathrm{AB}}\) in the direction \(\overrightarrow{\mathrm{AB}}\)

= PQ. e = \(\left(x_2-x_1, y_2-y_1, z_2-z_1\right)(l, m, n)\)

= \(l\left(x_2-x_1\right)+m\left(y_2-y_1\right)+n\left(z_2-z_1\right)\)

Coordinates Angles Between Two Lines

Theorem.8. If l₁, m₁, n₁ and l₂, m₂, n₂ are d.cs. of two lines L₁, L₂ then an angle θ between them is given by cosθ  = l₁l₂ + m₁m₂ + n₁n₂

Proof. Let P = (l₁, m₁, n₁) and Q = (l₂, m₂, n₂) and

⇒  \(\overline{\mathrm{OP}}=\bar{e}_1=\left(l_1, m_1, n_1\right) \overline{\mathrm{OQ}}=\bar{e}_2=\left(l_2, m_2, n_2\right)\)

∴ \(\overleftrightarrow{\mathrm{OP}} \| \mathrm{L}_1 \text { and } \overleftrightarrow{\mathrm{OQ}} \| \mathrm{L}_2\)

Since θ is one of the angles between L₁, L₂, we take \((\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}})=\theta\)

∴ cos θ = \(\cos (\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}})\) = cos (OP, OQ) = cos(e₁, e₂)

= e₁.e₂ = (l₁, m₁, n₁).(l₂, m₂, n₂) = l₁l₂ + m1m₂ + n1n₂

∴ Angles between L₁, L₂ are θ, 180° – θ.

Note.1. \(\sin \theta=\left|\bar{e}_1 \times \bar{e}_2\right|=\left|\left(m_1 n_2-m_2 n_1, n_1 l_2-n_2 l_1, l_1 m_2-l_2 m_1\right)\right|\)

= \(\sqrt{\left[\left(m_1 n_2-m_2 n_1\right)^2+\left(n_1 l_2-n_2 l_1\right)^2+\left(l_1 m_2-l_2 m_1\right)^2\right]}=\sqrt{\left[\sum\left(m_1 n_2-m_2 n_1\right)^2\right]}\)

OR: \(\sin ^2 \theta=1-\cos ^2 \theta=\left(l_1^2+m_1^2+n_1^2\right)\left(l_2^2+m_2^2+n_2^2\right)-\cos ^2 \theta\)

= \(\left(l_1^2+m_1^2+n_1^2\right)\left(l_2^2+m_2^2+n_2^2\right)-\left(l_1 l_2+m_1 m_2+n_1 n_2\right)^2\)

= \(\left(m_1 n_2-m_2 n_1\right)^2+\left(n_1 l_2-n_2 l_1\right)^2+\left(l_1 m_2-l_2 m_1\right)^2\)

Coordinates Lagrange’s Identity

For any real numbers l₁, m₁, n₁, l₂, m₂, n₂

(l₁² + m₁² + n₁²)² + (n₁l₂ – n₂l₁)² + (l₁m₂ – l₂m₁)²

By simplifying L.H.S and regrouping the terms we can show that L.H.S = R.H.S.

Coordinates Solved problems

Example.1. Calculate the cosine of the angle A of the triangle with vertices a(1, -1, 2), B(6, 11, 2), C(1, 2, 6).

Solution.

Given A = (1,-1,2), B – (6,11,2), C = (1,2,6)

We have \(\overline{\mathrm{AB}}=(6-1,11+1,2-2) \text { and } \overline{\mathrm{AC}}=(1-1,2+1,6-2)\)

cos \(\mathrm{A}=\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{AC}}}{|\overline{\mathrm{AB}}||\overline{\mathrm{AC}}|}=\frac{(5,12,0) \cdot(0,3,4)}{\sqrt{25+144+0} \cdot \sqrt{0+9+16}}=\frac{0+36+0}{13 \times 5}=\frac{36}{65}\)

OR: D.rs. of \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}\) are (6-1, 11+1, 2+2) and

(1-1,2+1,6-2) i.e., (5,12,0), (0,3,4)

∴ D.cs. of \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}} \text { are }\left(\frac{5}{13}, \frac{12}{13}, 0\right),\left(0, \frac{3}{5}, \frac{4}{5}\right) \text {. }\)

∴ \(\cos A=\frac{5}{13} \cdot 0+\frac{12}{13} \cdot \frac{3}{5}+0 \cdot \frac{4}{5}=\frac{36}{65} .\)

Example.2. If (l₁, m₁, n₁), (l₂, m₂, n₂), (l₃, m₃,  n₃) are the d.cs. of three mutually perpendicular rays, then find the d.cs. of a ray whose d.rs. are l₁ + l₂ + l₃, m₁ + m₂ + m₃, n₁ + n₂ + n₃. Hence, it shows that the ray is equally inclined to the given rays.

Solution.

Given

If (l₁, m₁, n₁), (l₂, m₂, n₂), (l₃, m₃,  n₃) are the d.cs. of three mutually perpendicular rays

Let L₁, L₂, L₃ be three mutually perpendicular rays whose d.cs. are (l₁,m₁,n₁), (l₂,m₂,n₂), (l₃,m₃,n₃)

∴ \(l_1^2+m_1^2+n_1^2=1, l_2^2+m_2^2+n_2^2=1, l_3^2+m_3^2+n_3^2=1 \text {, }\)

⇒ \(l_1 l_2+m_1 m_2+n_1 n_2=0, l_2 l_3+m_2 m_3+n_2 n_3=0, l_3 l_1+m_3 m_1+n_3 n_1=0\)

Now \(\left(l_1+l_2+l_3\right)^2+\left(m_1+m_2+m_3\right)^2+\left(n_1+n_2+n_3\right)^2\)

= \(\left(l_1^2+m_1^2+n_1^2\right)+\left(l_2^2+m_2^2+n_2^2\right)+\left(l_3^2+m_3^2+n_3^2\right)\)

+ \(2\left(l_1 l_2+m_1 m_2+n_1 n_2\right)+2\left(l_2 l_3+m_2 m_3+n_2 n_3\right)+2\left(l_3 l_1+m_3 m_1+n_3 n_1\right)=3\)

∴ d.cs. of the ray L whose d.rs. are \(l_1+l_2+l_3, m_1+m_2+m_3 \text {, are } n_1+n_2+n_3\)

are \(\frac{l_1+l_2+l_3}{\sqrt{3}}, \frac{m_1+m_2+m_3}{\sqrt{3}}, \frac{n_1+n_2+n_3}{\sqrt{3}}\)

If \(\left(\mathrm{L}, \mathrm{L}_1\right)=\theta \text {, then } \cos \theta=\frac{l_1\left(l_3+l_2+l_3\right)+m_1\left(m_1+m_2+m_3\right)+n_1\left(n_1+n_2+n_3\right)}{\sqrt{3}}=\frac{1}{\sqrt{3}}\)

i.e., \(\theta={Cos}^{-1}(1 / \sqrt{3})\). Similarly, we can have \(\left(\mathrm{L}, \mathrm{L}_2\right)=\left(\mathrm{L}, l_3\right)={Cos}^{-1}(1 / \sqrt{3}) \text {. }\)

∴ L is equally inclined with L₁, L₂, L₃.

Example.3. L₁, L₂, L₃ are three concurrent rays whose d.cs. are (l₁, m₁, n₁), (l₂, m₂, n₂), (l₃, m₃,  n₃) respectively. Prove that L₁, L₂, L₃ are coplanar <=> \(\left|\begin{array}{lll}
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2 \\
l_3 & m_3 & n_3
\end{array}\right|=0\)

Solution.

Given

L₁, L₂, L₃ are three concurrent rays whose d.cs. are (l₁, m₁, n₁), (l₂, m₂, n₂), (l₃, m₃,  n₃) respectively.

Rays L₁, L₂, L₃ are concurrent and unit vectors along L₁, L₂, L₃ are

⇒ \(l_1 \bar{i}+m_1 \bar{j}+n_1 \bar{k}, l_2 \bar{i}+m_2 \bar{j}+n_2 \bar{k}, l_3 \bar{i}+m_3 \bar{j}+n_3 \bar{k} \text { i.e. }\left(l_1, m_1, n_1\right),\left(l_2, m_2, n_2\right),\left(l_3, m_3, n_3\right) \text {. }\)

L₁, L₂, L₃ are coplanar ⇔ \(\left[\left(l_1, m_1, n_1\right),\left(l_2, m_2, n_2\right),\left(l_3, m_3, n_3\right)\right]=0 \Leftrightarrow\left|\begin{array}{lll}
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2 \\
l_3 & m_3 & n_3
\end{array}\right|=0\)

Example.4. Find the foot of the perpendicular from p(1, 8, 4) to the line \(\overleftrightarrow{\mathrm{AB}}\) where  A = (0, -11, 4), B = (2, -3, 1).

Solution. Let (Q; A, B) = 1 : λ (1+λ≠0)

∴ Q = \(\left(\frac{2}{1+\lambda}, \frac{-3-11 \lambda}{1+\lambda}, \frac{1+4 \lambda}{1+\lambda}\right)\)

⇒ \(\overline{\mathrm{PQ}}=\left(\frac{2}{1+\lambda}-1, \frac{-3-11 \lambda}{1+\lambda}-8, \frac{1+4 \lambda}{1+\lambda}-4\right) \text { and } \overline{\mathrm{AB}}=(2,-3+11,1-4)=(2,8,-3)\)

If \(\mathrm{PQ} \perp \mathrm{AB} \text {, then } \overline{\mathrm{AB}} \cdot \overline{\mathrm{PQ}}=0\)

∴ \(2\left(\frac{2-1-\lambda}{1+\lambda}\right)+8\left(\frac{-3-11 \lambda-8-8 \lambda}{1+\lambda}\right)-3\left(\frac{1+4 \lambda-4-4 \lambda}{1+\lambda}\right)=0\)

⇒ \(2-2 \lambda-88-152 \lambda+9=0 \Rightarrow 154 \lambda=-77 \Rightarrow \lambda=-\frac{1}{2} \Rightarrow 1: \lambda=1:-\frac{1}{2}\)

∴ Foot of the perpendicular from P to \(\overleftrightarrow{\mathrm{AB}}=\left(\frac{2}{(1 / 2)}, \frac{-3+\frac{11}{2}}{(1 / 2)}, \frac{1-2}{(1 / 2)}\right)=(4,5,-2)\)

Note. Length of the perpendicular from P to the line

∴ \(\overrightarrow{\mathrm{AB}}=\sqrt{(4-1)^2+(5-8)^2+(-2-4)^2}=3 \sqrt{6}\)

Example.5. If P, Q, R, S are the points (-1, 2, 4), (1, 0, 5), (3, 4, 5), (4, 6, 3), find the projection of \(\overline{\mathrm{PQ}} \text { on } \overleftrightarrow{R S}\) in the direction of \(\overline{R S}\)

Solution.

Given

If P, Q, R, S are the points (-1, 2, 4), (1, 0, 5), (3, 4, 5), (4, 6, 3)

⇒ \(\overline{\mathrm{PQ}}=(2,-2,1) \text { and } \overline{\mathrm{RS}}=(1,2,-2)\)

∴ Projection of \(\overline{\mathrm{PQ}} \text { on } \overrightarrow{\mathrm{RS}}\) in the direction of \(\overline{\mathrm{RS}}\)

= \(\frac{\overline{\mathrm{PQ}} \cdot \overline{\mathrm{RS}}}{|\overline{\mathrm{RS}}|}=\frac{(2,-2,1),(1,2,-2)}{|(1,2,-2)|}=\frac{(2) 1+(-2) 2+1(-2)}{\sqrt{[1+4+4]}}=\frac{-4}{3}\)

Example.6. Find the area of the △OAB where O is the origin, A = (x₁, y₁, z₁) and B = (x₂, y₂, z₂).

Solution.

Given

O is the origin, A = (x₁, y₁, z₁) and B = (x₂, y₂, z₂)

⇒ \(\overline{\mathrm{OA}}=\left(x_1, y_1, z_1\right) \text { and } \overline{\mathrm{OB}}=\left(x_2, y_2, z_2\right) \text {. }\)

Area of the \(\Delta \mathrm{OAB}=\frac{1}{2}|\overline{\mathrm{OA}} \times \overline{\mathrm{OB}}|\)

But \(\overline{\mathrm{OA}} \times \overline{\mathrm{OB}}=\left|\begin{array}{ccc}
\bar{l} & \bar{j} & \bar{k} \\
x_1 & y_1 & z_1 \\
x_2 & y_2 & z_2
\end{array}\right|=\left(y_1 z_2-y_2 z_1, z_1 x_2-x_1 z_2, x_1 y_2-x_2 y_1\right)\)

∴ Area of \(\Delta \mathrm{OAB}=\frac{1}{2} \sqrt{\left\{\left(y_1 z_2-y_2 z_1\right)^2+\left(z_1 x_2-z_2 x_1\right)^2+\left(x_1 y_2-x_2 y_1\right)^2\right\}} \text { sq. units }\)

Right Line Representation Of Line

Definition Of A Right Line In Geometry With Examples

1. Consider XZ and XY planes. Their common line of intersection is X-axis.

p(x, y, z) ∈ X-axis <=> P ∈ XY plane and P ∈ XZ plane <=> z = 0 and y = 0

∴ Equations to the line x-axis are y = 0, z = 0 i.e., equations to the plane of the plane passing through the X-axis.

Similarly, equations to y-axis are x = 0, z = 0 i.e., equations to the planes through the y-axis, and equations to z-axis are y = 0, x = 0 i.e., equations to the planes through the z-axis.

2. Consider any line L and two planes π₁, π₂ whose line of intersection is L.

Let the equations to π₁, π₂ be respectively

a₁x + b₁y + c₁z + d₁ = 0 …..(1) a₂x + b₂y + c₂z + d₂ = 0 …..(2)

P(x, y, z) ∈ L <=> P ∈ π₁ and P ∈ π₂

<=> a₁x + b₁y + c₁z + d₁ = 0, a₂x + b₂y + c₂z + d₂ = 0

∴ Equations to the line L are a₁x + b₁y + c₁z + d₁ = 0, a₂x + b₂y + c₂z + d₂ = 0

Thus: A line is represented by the equations of two planes through the line. Since any pair of planes can be taken through the line, the pairs of equations of the line are infinitely many.

Right Line Parametric Form

Theorem.1. Equations to the line passing through the point (x₁, y₁, z₁) and having d.cs. l, m, n are x = x₁ + lr, y = y₁ + mr, z = z₁ + nr, r being any real number.

Proof.

Let L be the required line and A (x₁, y₁, z₁). d.cs. of L are l,m,n.

Let P = (x,y,z) ∈ L. Let AP = |r|.

The projection of AP on the x-axis = x – x₁ = lr

Similarly y – y₁ = mr, z – z₁ = nr

⇒ x = x₁ + lr, y = y₁ + mr, z = z₁ + nr r being any real number.

∴ Equations to L are x = x₁ + lr, y = y + mr, z = z₁ + nr

i.e., \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\)

Note: 1. \(\frac{a_1}{l_1}=\frac{a_2}{l_2}=\frac{a_3}{l_3} \Leftrightarrow a_1: l_1=a_2: l_2=a_3: l_3\) and if any of l’s is zero, the corresponding a is also zero.

2. Equations of the line in the parametric form x = x₁ + lr, y = y₁ + mr, z = z₁ + nr can be written as

⇒ \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}(=r)\)

This form of equations of the line is called the equations of the line in the ‘Symmetric form’.

Let m = 0 = n, l ≠ 0, m ≠ 0. Equations to the line L are y – y₁ = 0, z – z₁ = 0. Then L represents a line parallel to the x-axis.

Theorem.2. Equations of a line through two distinct points (x₁, y₁, z₁) and (x₂, y₂, z₂) are \(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} \text { or } \frac{x-x_2}{x_2-x_1}=\frac{y-y_2}{y_2-y_1}=\frac{z-z_2}{z_2-z_1}\)

Proof.

Let L be the required line. Since (x₁,  y₁, z₁), (x₂, y₂, z₂) ∈ L, d.rs. of L are x₂ – x₁, y₂ – y₁, z₂ – z₁.

∴ Equations to L are \(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1} \text { or } \frac{x-x_2}{x_2-x_1}=\frac{y-y_2}{y_2-y_1}=\frac{z-z_2}{z_2-z_1}\)

Theorems Related To Right Lines In Mathematics

Theorem.3. Transform the equations a₁x + b₁y + c₁z + d₁ = 0 a₂x + b2 + cz₂ + d₂ = 0 of the line of symmetrical form.

Proof.

Let L be the line of intersection of the planes

a₁x + b₁y + c₁z + d₁ = 0        …(1)

a₂x + b₂y + c₂z + d₂ = 0       …(2)

Let [l,m,n] be the D.cs. of L.

L lies in both the places (1) and (2).

Since the d.rs. of the normals to the planes are (a₁,b₁,c₁) and (a₂,b₂,c₂) we have a₁l + b₁m + c₁n = 0, a₂l + b₂m + c₂n = 0

⇒ \(\frac{l}{b_1 c_2-b_2 c_1}=\frac{m}{c_1 a_2-c_2 a_1}=\frac{n}{a_1 b_2-a_2 b_1}\)

Without loss of generality, we can take a₁b₂ – a₂b₁ ≠ 0.

Now to find the equations to L, we require a point on L.

Let L intersect, say, the XY plane i.e. Z = 0 at P.

∴ From (1) and (2): a₁x + b₁y + d₁ = 0, a₂x + b₂y + d₂ = 0.

Solving: \(x=\frac{b_1 d_2-b_2 d_1}{a_1 b_2-a_2 b_1}, y=\frac{d_1 a_2-d_2 a_1}{a_1 b_2-a_2 b_1}\)

∴ \(\mathrm{P}=\left(\frac{b_1 d_2-b_2 d_1}{a_1 b_2-a_2 b_1}, \frac{d_1 a_2-d_2 a_1}{a_1 b_2-a_2 b_1}, 0\right)\) is a point on L.

∴ Equations to L in the symmetric form are

⇒ \(\frac{x-\frac{b_1 d_2-b_2 d_1}{a_1 b_2-a_2 b_1}}{b_1 c_2-b_2 c_1}=\frac{y-\frac{d_1 a_2-d_2 a_1}{a_1 b_2-a_2 b_1}}{c_1 a_2-c_2 a_1}=\frac{z-0}{a_1 b_2-a_2 b_1}\)

Example. Write the equations of the line x = ay + b and z = cy + d in the symmetrical form.

Solution. 

Given

x = ay + b and z = cy + d

Equations of the line L are lx + (-a)y + 0. z = b and ox + cy + (-1)z = -d.

Let d.rs. of the line L be l,m,n.

∴ 1.l + (-a)m + 0(n) = 0, 0.l + cm + (-l)n = 0 i.e., \(\frac{l}{a}=\frac{m}{l}=\frac{n}{c}\)

A point on L is (b,0,d).

∴ L is \(\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c}\)

OR: Equations to the line are x = ax + b, z = cy + d.

They can be written as \(\frac{x-b}{a}=\frac{y}{1}, \frac{z-d}{c}=\frac{y}{1} \text {. i.e. } \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}\) which form is the symmetrical form of the equations of the line.

Note. If the equations of a line are given as a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0, then the equations of the line are said to be in unsymmetrical form.

Right Line Solved Problems

Example 1. Find the distance of the point (1, -2, 3) from the plane x – y + z = 5 measured parallel to the line whose d.cs. are proportional to 2, 3, -6.

Solution.

Given

The point (1, -2, 3) and the plane x – y + z = 5

Let L be the line through the point P(1, -2, 3) and parallel to the line with d.rs. 2, 3, -6.

∴ Equations to L are \(\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}\) (= t say)

Any point on L is Q(2t + 1, 3t – 2, -6t + 3). Let π be the plane x – y + z = 5.

Q ∈ π ⇒ 2t + 1 – 3t + 2 – 6t + 3 = 5 ⇒ -7t = 1 ⇒ t = (1/7)

∴ \(\mathrm{Q}=\left(\frac{9}{7}, \frac{-11}{7}, \frac{15}{7}\right)\)  ∴\( \mathrm{PQ}^2=\left(\frac{9}{7}-1\right)^2+\left(\frac{-11}{7}+2\right)^2+\left(\frac{15}{7}-3\right)^2=\frac{4+9+36}{49}=1\)

∴ Required distance = 1

Example 2. Find the image of the point (2, -1, 3) in the plane 3x – 2y + z = 9.

Solution.

Given

The point (2, -1, 3) and the plane 3x – 2y + z = 9

Let P = (2,-1,3). Let π be the plane 3x – 2y + z = 9.

Let Q = (x₁, y₁, z₁) be the image of P in π.

∴ \(\overleftrightarrow{\mathrm{PQ}} \perp \pi\) ∴ Drs. of \(\overleftrightarrow{\mathrm{PQ}}\) are 3, -2, 1.

∴ Equation to \(\overleftrightarrow{\mathrm{PQ}}\) are \(\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-3}{1}\) (= t say)

Let R be the midpoint of PQ.

Let R = (3t + 2, -2t -1, t + 3)

R ∈ π ⇒ 9t + 6 + 4t + 2 + t + 3 = 9 ⇒ t = -/frac{1}{7}

∴ R = \(\left(\frac{-3}{7}+2, \frac{2}{7}-1, \frac{-1}{7}+3\right)=\left(\frac{11}{7}, \frac{-5}{7}, \frac{20}{7}\right)\)

R is the midpoint of PQ

⇒ \(2+x_1=\frac{22}{7} \text { i.e., } x_1=\frac{8}{7} ; \quad-1+y_1=\frac{-10}{7}\)

⇒ \(\text { i.e., } y_1=\frac{-3}{7} ; \quad 3+z_1=\frac{40}{7} \text { i.e., } z_1=\frac{19}{7}\)

∴ Q = Image of P in π = \(\left(\frac{8}{7}, \frac{-3}{7}, \frac{19}{7}\right)\)

Solved Exercise Problems On Right Lines Step-By-Step

Example 3. Find the foot of the perpendicular from (1, 2, 3) to the plane x + 2y + 3z + 4 = 0

Solution.

Let the given plane is π = x + 2y + 3z + 4 = 0 …(1)

Given point is P = (1,2,3)

Let the foot of the perpendicular from P to the plane is \(\mathrm{Q}=\left(x_1, y_1, z_1\right)\)

Equations of \(\overline{\mathrm{PQ}}, \frac{x_1-1}{1}=\frac{y_1-2}{2}=\frac{z_1-3}{3}=t \text { (say) }\)

∴ \(x_1=t+1, y_1=2 t+2, z_1=3 t+3\)

Since Q lies on the plane, we have

(t + 1) + 2(2t + 2) + 3(3t + 3) + 4 = 0

⇒ \(14 t+18=0 \Rightarrow t=\frac{-9}{7}\)

⇒ \(x_1=\frac{-9}{7}+1=\frac{-2}{7}, y_1=2 t+2=\frac{-18}{7}+2=-\frac{4}{7}, z_1=3 t+3=\frac{-27}{7}+3=\frac{-6}{7}\)

∴ Foot of the perpendicular = P = \(\left(x_1, y_1, z_1\right)=\left(\frac{-2}{7}, \frac{-4}{7}, \frac{-6}{7}\right)\)

Example 4. Find the angles between the lines x – 2y + z = 0, x + y – z = 3;……L₁ x + 2y + z = 5, 8x + 12y + 5z = 0……L₂.

Solution.

Let the planes be

x – 2y + z = 0       …(1)

x + y – z = 3        …(2)

x + 2y + z = 5    …(3)

8x + 12y + 5z = 0     …(4)

(1), (2) represent the line L₁ and (3), (4) represent the line L₂.

The d.rs. of the normals to the planes (1), (2), (3), (4) are (1, -2, 1), (1, 1, -1) (1, 2, 1) and (8, 12, 5) respectively.

If \(\left(I_1, m_1, n_1\right) \text { and }\left(l_2, m_2, n_2\right)\) and the d.rs. of the lines L₁ and L₂ respectively, \(\left(l_1, m_1, n_1\right)=(2-1,1+1,1+2)=(1,2,3) \text {; }\)

⇒ \(\left(l_2, m_2, n_2\right)=(10-12,8-5,12-16)=(-2,3,-4)\)

∴ D.rs. of L₁ are 1, 2, 3, and d.rs. of L₂ are -2, 3, -4

If θ is one of the angles between L₁, L₂ then

⇒ \(\cos \theta=\frac{-2+6-12}{\sqrt{1+4+9} \cdot \sqrt{4+9+16}}=\frac{-8}{\sqrt{406}}\)

∴ \(\left(L_1, L_2\right)={Cos}^{-1}\left(\frac{-8}{\sqrt{(406)}}\right), \pi-{Cos}^{-1}\left(\frac{-8}{\sqrt{(406)}}\right)\)

Example 5. Find the equations to the line L₁ passing through the origin and perpendicular to the L₂ whose equations are \(\frac{x-2}{1}=\frac{y+3}{-2}=\frac{z}{1}\). Also find the foot of the perpendicular from the origin to L₂.

Solution.

Equations to L₂ are \(\frac{x-2}{1}=\frac{y+3}{-2}=\frac{z}{1} \quad(=t \text { say })\)

Any point L₂ is (t+2, -2t-3, t).

Let P be the foot of the perpendicular from the origin to the line L₂.

If P = (t+2, -2t-3,t) then d.rs. of (∵ \(\mathrm{L}_{\mathrm{l}}=\overleftrightarrow{\mathrm{OP}}\)) are t + 2, -2t -3, t.

⇒ \(\mathrm{L}_1 \perp \mathrm{L}_2 \Rightarrow 1(t+2)-2(-2 t-3)+1 t=0 \Rightarrow 6 t=-8 \Rightarrow t=-\frac{4}{3}\)

∴ A foot of the perpendicular from the origin to L₂

= \(\left(\frac{-4}{3}+2, \frac{8}{3}-3, \frac{-4}{3}\right)=\left(\frac{2}{3}, \frac{-1}{3}, \frac{-4}{3}\right)\)

∴ Equations to L1 are \(\frac{x-0}{(2 / 3)}=\frac{y-0}{(-1 / 3)}=\frac{z-0}{(-4 / 3)} \text { i.e., } \frac{x}{2}=\frac{-y}{1}=\frac{-z}{4}\)

Properties Of Right Lines With Examples And Solutions 

Example 6. A variable plane makes intercepts on the axes, the sum of whose squares is k² (a constant). Show that the locus of the foot of the perpendicular from origin to the plane is (x^-2 + y^-2 + z^-2)(x² + y² + z²)² = k²

Solution.

Given

A variable plane makes intercepts on the axes, the sum of whose squares is k² (a constant)

π is a variable plane \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \quad(a b c \neq 0)\)

⇒ intercepts of π on the axes are a, b, c ⇒ \(a^2+b^2+c^2=k^2 \text { (given) }\)          …(1)

D.rs. of normal to π are \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\)

Let L be the line through O and perpendicular to π.

∴ Equations to L are \(\frac{x}{1 / a}=\frac{y}{1 / b}=\frac{z}{1 / c}(=t \text { say }) .\)

∴ Any point P on L is \(\left(\frac{t}{a}, \frac{t}{b}, \frac{t}{c}\right) .\)

P is the foot of the perpendicular from O to π

⇒ P ∈ π ⇒ \(\frac{t}{a^2}+\frac{t}{b^2}+\frac{t}{c^2}=1 \Rightarrow t=\left(a^{-2}+b^{-2}+c^{-2}\right)^{-1}\)

∴ \(\mathbf{P}=\left(a^{-1}\left(a^{-2}+b^{-2}+c^{-2}\right)^{-1}, b^{-1}\left(a^{-2}+b^{-2}+c^{-2}\right)^{-1}, c^{-1}\left(a^{-2}+b^{-2}+c^{-2}\right)^{-1}\right) \text {. }\)

P = \(\left(x_1, y_1, z_1\right) \Rightarrow x_1=a^{-1}\left(a^{-2}+b^{-2}+c^{-2}\right)^{-1} ; y_1=b^{-1}\left(a^{-2}+b^{-2}+c^{-2}\right)^{-1} \text {; }\)

⇒ \(z_1=c^{-1}\left(a^{-2}+b^{-2}+c^{-2}\right)^{-1} \text {. }\)

⇒ \(x_1^2+y_1^2+z_1^2=\left[\left(a^{-2}+b^{-2}+c^{-2}\right)^{-1}\right]^2\left(a^{-2}+b^{-2}+c^{-2}\right)=\left(a^{-2}+b^{-2}+c^{-2}\right)^{-1} \text { and }\)

⇒ \(x_1^{-2}+y_1^{-2}+z_1^{-2}=\left(a^{-2}+b^{-2}+c^{-2}\right)^2\left(a^2+b^2+c^2\right)\)

⇒ \(\left(x_1^{-2}+y_1^{-2}+z_1^{-2}\right)=\frac{1}{\left(x_1^2+y_1^2+z_1^2\right)^2} \cdot k^2\)      [using (1)]

⇒ \(\left(x_1^{-2}+y_1^{-2}+z_1^{-2}\right)\left(x_1^2+y_1^2+z_1^2\right)^2=k^2\)

∴ Locus of P is \(\left(x^{-2}+y^{-2}+z^{-2}\right)\left(x^2+y^2+z^2\right)^2=k^2\)

Example 7. The plane lx + my + nz = p, l² + m² + n² = 1, p > 0 meets the axes in P, Q, R and G is the centroid of the △PQR, If the perpendicular line to the plane at G meets the coordinate planes in A, B, C then prove that \(\frac{1}{\mathrm{GA}}+\frac{1}{\mathrm{~GB}}+\frac{1}{\mathrm{GC}}=\frac{3}{p}\)

Solution.

Given

The plane lx + my + nz = p, l² + m² + n² = 1, p > 0 meets the axes in P, Q, R and G is the centroid of the △PQR, If the perpendicular line to the plane at G meets the coordinate planes in A, B, C

Let π be the plane lx + my + nz = p, \(l^2+m^2+n^2=1\).

π meets the axes in P, Q, R.

∴ \(\mathrm{P}=\left(\frac{p}{l}, 0,0\right), \mathrm{Q}=\left(0, \frac{p}{m}, 0\right), \mathrm{R}=\left(0,0, \frac{p}{n}\right)\)

∴ Centroid G of △PQR = (p/3l, p/3m, p/3n)

Let L be the line perpendicular to π at G.

∴ Equations to L are \(\frac{x-(p / 3 l)}{l}=\frac{y-(p / 3 m)}{m}=\frac{z-(p / 3 n)}{n}\) (=t say)

L meets the YZ plane i.e. x = 0 in A.

∴ \(|t|=\mathrm{GA}=\left|\frac{0-(p / 3 l)}{l}\right|=\left|\frac{-p}{3 l^2}\right| \text { i.e., } \frac{1}{\mathrm{GA}}=\frac{3 l^2}{p}(p>0)\)

Similarly \(\frac{1}{\mathrm{~GB}}=\frac{3 m^2}{p} \text { and } \frac{1}{\mathrm{GC}}=\frac{3 n^2}{p}\)

∴ \(\frac{1}{\mathrm{GA}}+\frac{1}{\mathrm{~GB}}+\frac{1}{\mathrm{GC}}=\frac{3 l^2+3 m^2+3 n^2}{p}=\frac{3}{p}\)

Right Line Angle Between A Line And A Plane

Theorem.4. If θ is the acute angle between the line \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\) and the plane ax + by + cz + d = 0, then \(\sin \theta=\pm \frac{a l+b m+c n}{\sqrt{a^2+b^2+c^2} \sqrt{l^2+m^2+n^2}}\)

Proof.

Let L be the given line and E  be the given plane.

The d.rs. of L are (l, m, n) and the d.rs. of the normal to E are (a, b, c).

Since θ is the acute angle between L and E, the angles between L and normal to E are 90° ± θ

⇒ \(\cos (90 \pm \theta)= \pm \sin \theta=\frac{a l+b m+c n}{\sqrt{a^2+b^2+c^2} \sqrt{l^2+m^2+n^2}}\)

⇒ \(\sin \theta= \pm \frac{a l+b m+c n}{\sqrt{a^2+b^2+c^2} \sqrt{l^2+m^2+n^2}}\)

Right Line Conditions For A Line To Lie In A Plane

Theorem.5. If L is the line \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\) and π is the plane ax + by + cz + d = 0, then L ⊂ π <=> a₁x + b₁y + c₁z + d = 0, al + bm + cn = 0.

Proof.

Any point P on L is \(\left(x_1+l r, y_1+m r, z_1+n r\right)\) where r is any real number.

L ⊂ π ⇔  P  ∈ π

⇔ \(a\left(x_1+l r\right)+b\left(y_1+m r\right)+c\left(z_1+n r\right)+d=0 \text { for any real number }\)

⇔ \(\left(a x_1+b y_1+c z_1+d\right)+r(a l+b m+c n)=0\) for any real number r

⇔ \(a x_1+b y_1+c z_1+d=0, a l+b m+c n=0\)

Note.1. A-line lies in a plane if

(1) any point on the line lies in the plane, and

(2) the normal to the plane is perpendicular to the line.

2. Equation to a plane containing the line \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\) can be taken as a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 where a, b, c are parameters such that al + bm + cn = 0.

3. The line \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\) is parallel to the plane ax + by + cz + d = 0 and not contained in the plane ⇒ al + bm + cn = 0 and ax₁ + by₁ + cz₁ ≠ 0.

4. The line \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\) is perpendicular to the plane ax + by + cz + d = 0 ⇒ \(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}\)

Right Line Solved Problems

Example.1. Find the equation to a plane through the line \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\) and parallel to another line with d.cs. l₂, m₂, n₂.

Solution. 

Equation to the plane through the line \(\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}\) can be taken as

⇒ \(l\left(x-x_1\right)+m\left(y-y_1\right)+n\left(z-z_1\right)=0\)         …(1)

where \(l l_1+m m_1+m n_1=0\)          …(2)

If (1) is parallel to another line with d.cs. l2, m2, n2

then \(l l_2+m m_2+n n_2=0\)       …(3)

∴ From (2) and (3), \(\frac{l}{m_1 n_2-m_2 n_1}=\frac{m}{n_1 l_2-n_2 l_1}=\frac{n}{l_1 m_2-l_2 m_1} \quad \text { (= Ksay) }\)

∴ The equation to the required plane is

⇒ \(\left(m_1 n_2-m_2 n_1\right)\left(x-x_1\right)+\left(n_1 l_2-n_2 l_1\right)\left(y-y_1\right)+\left(l_1 m_2-l_2 m_1\right)\left(z-z_1\right)=0\)

i.e. \(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{array}\right|=0\)

(may also be obtained by eliminating l, m, n from (1), (2), (3))

Step-By-Step Guide To Solving Right Line Problems In Geometry

Example.2. Find the equation to the plane through the line \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\) and through the point (x₁, y₁, z₁)

Solution.

Let L be the line \(\frac{x-x_2}{l}=\frac{y-y_2}{m}=\frac{z-z_2}{n} .\)

∴ L passes through the point (x₂, y₂, z₂).

Let π be the plane containing the line L and passing through the point (x₁, y₁, z₁).

Let the equation to the plane π be \(a\left(x-x_2\right)+b\left(y-y_2\right)+c\left(z-z_2\right)=0\)       …(1)

where al + bm + cn = 0          …(2)

Also \(a\left(x_1-x_2\right)+b\left(y_1-y_2\right)+c\left(z_1-z_2\right)=0\)       …(3)

Eliminating a, b, c from (1), (3), and (2), the equation to the plane π is

⇒ \(\left|\begin{array}{ccc}
x-x_2 & y-y_2 & z-z_2 \\
x_1-x_2 & y_1-y_2 & z_1-z_2 \\
l & m & n
\end{array}\right|=0\)

Example.3. Find the equation of the plane through the origin and containing the line x – 3y + 2z + 3 = 0 = 3x – y + 2z – 5.

Solution.

Given line is x – 3y + 2z + 3 = 0 = 3x – y + 2z – 5.

Any plane through the given line is x – 3y + 2z + 3 + λ(3x – y + 2z – 5) = 0

If it passes through the origin, then 0 + 3 + λ(0-5) = 0 ⇒ λ = 3/5

∴ Equation to the required plane is x – 3y + 2z + 3 + (3/5)(3x – y + 2z –  5) = 0 i.e. 7x – 9y + 8z = 0.

Example.4. Find the equation of the plane that passes through the line a₁x + b₁y + c₁z + d₁ = 0 = a₂x + b₂y + c₂z + d₂ and is parallel to the line \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n}\)

Solution.

Let the plane through the line \(a_1 x+b_1 y+c_1 z+d_1=0=a_2 x+b_2 y+c_2 z+d_2\) and parallel to \(\frac{x}{l}=\frac{y}{m}=\frac{z}{n} \text { be } a_1 x+b_1 y+c_1 z+d_1+\lambda\left(a_2 x+b_2 y+c_2 z+d_2\right)=0\) where

i.e., \(\lambda=-\frac{l a_1+m b_1+n c_1}{l a_2+m b_2+n c_2},\left(l a_2+m b_2+n c^2 \neq 0\right)\)

∴ The equation to the required plane is

⇒ \(\left(l a_2+m b_2+n c_2\right)\left(a_1 x+b_1 y+c_1 z+d_1\right)-\left(l a_1+m b_1+n c_1\right)\left(a_2 x+b_2 y+c_2 z+d_2\right)=0\)

Examples Of Right Line Problems In 2d And 3d Geometry

Example.5. A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c) and P = (a, b, c) are points distinct from O such that P² = a² + b² + c² (p > 0) and q^-2 = a^-2 + b^-2 + c^-2 (q > 0). If θ is the acute angle between \(\overleftrightarrow{\mathrm{OP}}, \overleftrightarrow{\mathrm{ABC}}\) show that \(\sin \theta=\frac{3 q}{p}\).

Solution.

Equation to the plane \(\overleftrightarrow{\mathrm{ABC}} \text { is } \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \text {. }\)

D.rs. of its normal are (1/a), (1/b), (1/c).

D.rs. of \(\overleftrightarrow{\mathrm{OP}}\) are a, b, c.

∴ \(\theta=(\overleftrightarrow{\mathrm{OP}}, \overleftrightarrow{\mathrm{ABC}})\)

⇒ \(\sin \theta=\frac{a \cdot \frac{1}{a}+b \cdot \frac{1}{b}+c \cdot \frac{1}{c}}{\sqrt{a^2+b^2+c^2} \cdot \sqrt{\left(1 / a^2\right)+\left(1 / b^2\right)+\left(1 / c^2\right)}} \Rightarrow \sin \theta=\frac{3}{p \cdot \frac{1}{q}}=\frac{3 q}{p}\)

Right Line Coplanarity Of Lines

Theorem.6. L₁, L₂ are lines whose equations are \(\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}\) …..(1)\(\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}\)….(2). L₁, L₂ are coplanar ⇒ \(\left|\begin{array}{ccc}
x_1-x_2 & y_1-y_2 & z_1-z_2 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{array}\right|=0\)

Proof:

First Method.

An equation to the plane containing line (1) is

⇒ \(a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0\)           …(3)

with the condition that \(a l_1+b m_1+c n_1=0\)       …(4)

where not all a, b, c are zero.

The line (2) lies on the plane (3) if (1) the point \(\left(x_2, y_2, z_2\right)\) lies on (3)

⇒ \(a\left(x_2-x_1\right)+b\left(y_2-y_1\right)+c\left(z_2-z_1\right)=0\)    …(5)

and (2) the line (2) is perpendicular to the normal to the plane (3)

⇒ a l_2+b m_2+c n_2=0         …(6)

The given lines (1) and (2) are coplanar if the three linear homogeneous equations (4), (5), (6) in a, b, c are consistent.

Eliminating a, b, c from the equations (4), (5), (6) we get

⇒ \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{array}\right|=0\)       …(7)

This is the condition for the lines (1) and (2) be coplanar.

Theorem.7. Equation to the plane containing the line L1 with equations \(\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}\) and parallel to the line L2 with equations \(\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}\) is \(\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
l_1 & m_1 & n_1 \\
l_2 & m_2 & n_2
\end{array}\right|=0\)

Proof.

The lines are \(\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}\)      …(1)

⇒ \(\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}\)       …(2)

the equation to the plane containing the line (1) is