Vector Calculus 4th Edition Chapter 1 Vectors
Page 47 Problem 1 Answer
Given: The plane containing the point (3,−1,2) and perpendicular to i−j+2k.
To find: An equation for the plane.E
valuate to get the final answer.
The equation of a plane containing a point P(x0,y0,z0) and a vector perpendicular to the plane given by n=Ai+Bj+Ck is given by the following equation A(x−x0)+B(y−y0)+C(z−z0)=0.
Here point is (3,−1,2) and vector perpendicular is i−j+2k.
Hence using the above equation we get the equation of plane as:
1(x−3)−1(y+1)+2(z−2)=0.
Hence, the required solution is x−y+2z=8.
Page 47 Problem 2 Answer
Given : The point (9,5,−1) and perpendicular to i−2k.
To find: Equation for the plane.
Evaluate to get the final answer.
The equation of a plane containing a point P(x0,y0,z0) and a vector perpendicular to the plane given by n=Ai+Bj+Ck is given by the following equation A(x−x0)+B(y−y0)+C(z−z0)=0.
Here point is (9,5,−1) and vector perpendicular is i−2k.
Hence using the above equation we get the equation of plane as:
1(x−9)−2(z+1)=0
Hence, the required solution is x−2z=11.
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Page 47 Problem 3 Answer
Given: Plane containing the points (A,0,0),(0,B,0), and (0,0,C)
at least two of A, B, and C are nonzero.
To find: An equation for the plane containing the given points.
Evaluate the expression to obtain the final answer.
Let’s apply cross product of two vectors on the plane.
(A−0,0−B,0−0)×(0−0,0−B,C−0)=(−BC,−AC,−AB)
Hence, above vector is perpendicular to the plane containing the three points.
Here, equation of a plane containing a point P(x0,y0, z0)
Vector perpendicular to the given plan is n=Ji+Kj+Lk
A(x−x0)+B(y−y0)+C(z−z0)=0
By taking points(A,0,0) and vector perpendicular is −BCi−ACj−ABk
Therefore , the equation is BC(x−A)−AC(y−0)−AB(z−0)=0
Hence, equation for the plane containing the given points is BCx+ACy+ABz=ABC
Page 47 Problem 4 Answer
Given: A plane 5x−4y+z=1 passes through a point (2,−1,−2)
To find: An equation for the plane that is parallel to the given plan.
Evaluate the expression to obtain the final answer
If two planes are parallel, then the vector perpendicular to the plane is also perpendicular to the desired plan
Therefore,
For a plan 5x−4y+z=1, perpendicular vector is (5,−4,1)
The above vector is perpendicular to the plane containing the point (2,−1,−2).
Here, the equation of a plane containing a point P(x0,y0, z0)
Vector perpendicular to the given plan is n=Ai+Bj+Ck
A(x−x0)+B(y−y0)+C(z−z0)=0
By taking points (2,−1,−2) and vector perpendicular is 5i−4j+k
Therefore, the equation is
5(x−2)−4(y+1)+(z+2)=0
Hence, 5x−4y+z=12 is an equation for the plane that is parallel to the given plan.
Page 47 Problem 5 Answer
Given: A plane 2x−3y+z=5 and that passes through the point (−1,1,2)
To find: An equation for the plane that is parallel to the given plan.
Evaluate the expression to obtain the final answer.
If two planes are parallel, then the vector perpendicular to the plane is also perpendicular to the desired plan
Therefore,
For a plan 2x−3y+z=5, perpendicular vector is (2,−3,1)
The above vector is perpendicular to the plane containing the point (−1,1,2)
Here, the equation of a plane containing a point P(x0,y0, z0)
Vector perpendicular to the given plan is n=Ai+Bj+Ck
A(x−x0)+B(y−y0)+C(z−z0)=0
By taking points (−1,1,2) and vector perpendicular is 2i−3j+k
Therefore, The equation is 2(x+1)−3(y−1)+(z−2)=0
Hence, 2x−3y+z=−3 is an equation for the plane that is parallel to the given plan.
Page 47 Problem 6 Answer
Given: A plane x−y+7z=10 and hat passes through the point(−2,0,1)
To Find: An equation for the plane that is parallel to the given plan.
Evaluate the expression to obtain the final answer.
If two planes are parallel, then the vector perpendicular to the plane is also perpendicular to the desired plan.
Therefore, For a plan x−y+7z=10, perpendicular vector is (1,−1,7)
The above vector is perpendicular to the plane containing the point (−2,0.1)
Here, equation of a plane containing a point P(x0,y0, z0)
Vector perpendicular to the given plan is n=Ai+Bj+Ck
A(x−x0)+B(y−y0)+C(z−z0)=0
By taking points (−2,0.1) and vector perpendicular is i−j+7k
Therefore, The equation is 1(x+2)−1(y−0)+7(z−1)=0
Hence, from the above steps x−y+7z=5 is an equation for the plane that is parallel to the given plan.
Page 47 Problem 7 Answer
Given: A plane 2x+2y+z=5 that contains the line with parametric equations x=2−t
y=2t+1
z=3−2t
To Find: An equation for the plane that is parallel to the given plan.
Evaluate the expression to obtain the final answer.
If two planes are parallel, then the vector perpendicular to the plane is also perpendicular to the desired plan.
Therefore, For a plan 2x+2y+z=5, perpendicular vector is (2,2,1)
The above vector is perpendicular to the plane containing the point is given in an equation format
that is, 2x+2y+z=D……….(1)
Here D is unknown.
From the given, x=2−t
y=2t+1
z=3−2t
substitute in equation 1 to find D i,e 2(2−t)+2(2t+1)+1(3−2t)=D
4−2t+4t+2+3−2t=D
D=9
Hence, 2x+2y+z=9 is an equation for the plane that is parallel to the given plan.
Page 47 Problem 8 Answer
Given: The plan 5x−3y+2z=10 that contains the line with parametric equations x=t+4
y=3t−2
z=5−2t
To find: Why there is no plane parallel to the given plane.
Evaluate the expression to obtain the final answer.
If two planes are parallel, then the vector perpendicular to the plane is also perpendicular to the desired plan.
Therefore, For a plan 5x−3y+2z=10, perpendicular vector is (5,−3,2)
The above vector is perpendicular to the plane containing the point given by in an equation format
that is,
5x−3y+2z=D
Here D is unknown.
From the given, x=t+4
y=3t−2
z=5−2t substitute in equation 1 to find D value.
Hence,
D=5(t+4)−3(3t−2)+2(5−2t)
D=5t+20−9t+6+10−4t
D=36−8t
But the value of D is not constant which suggests that the line will surely intersect the planes parallel to 5x−3y+2z=10
Hence, it concludes that there is no plane parallel to the given plane which can completely contain the line.
Page 47 Problem 9 Answer
Given: The line x=3t−5
y=7−2t
z=8−t
and that passes through the point(1,−1,2).
To find: Find an equation for the plane that is perpendicular to the line.
Evaluate the expression to obtain the final answer.
Assume that,
The vector 3i−2j−k is parallel to the line
Hence, perpendicular to the plan which passes through (1,−1,2)
Therefore, the perpendicular to that vector is 3(x−1)−2(y+1)−(z−2)=0
Hence, 3(x−1)−2(y+1)−(z−2)=0 is an equation for the plane that is perpendicular to the line.
Page 47 Problem 10 Answer
Given: Equation of two lines l1:x=t+2,y=3t−5,z=5t+1,l2=x=5−t,y=3t−10,z=9−2t.
To Find: Equation of the plane that contains the given lines.
The vector i+3j+5k is parallel to the first line and the vector −i+3j−2k is parallel to the second line.
So the cross product of these two is perpendicular to the plane containing these two lines:
n=(i+3j+5k)×(−i+3j−2k)
=−21i−3j+6j
Now, find a single point on either line and it is known that point will also be in the plane.
Notice that (2,−5,1) is a point in the first line. So the equation of the plane is then
−21(x−2)−3(y+5)+6(z−1)=0
⇒7(x−2)+(y+5)−2(z−1)=0
Hence, the required equation of the plane is 7(x−2)+(y+5)−2(z−1)=0.
Page 47 Problem 11 Answer
Given: Equation of two planes x+2y−3z=5,5x+5y−z=1.
To Find: Line of equation of the planes.
A normal vector for the first plane is i+2j−3k and a normal vector for the second plane is 5i+5j−k.
A nonzero vector perpendicular to these will be parallel to the line of intersection.
So, find the cross product as follows:
(i+2j−3k)×(5i+5j−k)=13i−14j−5k
Now, find a point on the line of intersection.
Let x=0 and then solve the resulting system of equations:
{2y−3z=5
5y−z=1
⇒ y =−2/13 and z=−23/13.
Following will be the scalar equations
{x=13t
y=−2/13−14t
z=−23/13−5t
Hence, the required set of parametric equations:⎧
{x=13t
y=−2/13−14t
z=−23/13−5t
Page 47 Problem 12 Answer
Given: Equation of plane 2x−3y+5z=−1.
To Find: Set of parametric equation for the line perpendicular to the plane.
The fact that the line is perpendicular to the plane 2x−3y+5z=−1
⇒it is parallel to the vector 2i−3j+5k.
So, using the point(5,0,6) on the line, it is obtained:
{x=5+2t
y=−3t
z=6+5t
Hence, the required set of parametric equation of lines:{x=5+2t
y=−3t
z=6+5t
Page 47 Problem 13 Answer
Given: Equations of parallel planes 8x−6y+9Az=6, Ax+y+2z=3
To Find: The value of A.
For the planes to be parallel, they must be perpendicular to the same vectors.
Now, the first plane is perpendicular to every scalar multiple of 8i−6j+9Ak and the second plane is perpendicular to every scalar multiple of Ai+j+2k
It is required to find a solution to 8i−6j+9Ak=K(Ai+j+2k)
The one we obtained should be equal to:
{8=AK
−6=K
9A=2K
The unique solution to this is A=−4/3, K=−6.
⇒A=−4/3
The value of A=−4/3