Vector Calculus 4th Edition Chapter 1 Vectors
Page 16 Problem 1 Answer
Given: The vector(2,4)
To write: The given vector by using the standard basis vectors for R2 and R3.
Evaluate to get the required solution.
Firstly, the standard notation in R2:
a=(a1,a2)=a1i+12j
In R3:a=(a1,a2,a3)=a1i+a2j+a3k
Given vector in standard notation is: (2,4)=2i+4j
Therefore, we can write the given vector as (2,4)=2i+4j.
Page 16 Problem 2 Answer
Given: The vector (9,−6)
To find:The given vector by using the standard basis vectors for R2 and R3.
Evaluate to get the final result.
The Standard notation in R2:
a=(a1,a2)=a1i+a2j
In R3: a=(a1,a2,a3)=a1i+a2j+a3k
Given vector in standard notation is:
(9,−6)=9i−6j
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Therefore, rewriting the given vector we get 9i−6j.
Page 16 Problem 3 Answer
Given : The vector(3,π,−7)
To find :The given vector by using the standard basis vectors for R2 and R3 .
Evaluate to get the final result.
Standard notation in R2:a=(a1,a2)=a1i+a2j
In R3:
a=(a1,a2,a3)=a1i+a2j+a3k
Given vector in standard notation is:
(3,π,−7)=3i+πj−7k
Therefore, rewriting the given vector we get 3i+πj−7k.
Page 16 Problem 4 Answer
Given : The vector(−1,2,5)
To write: The given vector by using the standard basis vectors for R2 and R3.
Evaluate to get the final result.
Standard notation in R2:
a=(a1,a2)=a1i+a2j
In R2 :a=(a1,a2,a3)=a1i+a2j+a3k
Given vector in standard notation is: (−1,2,5)=−i+2j+5k
Therefore, we can write the given vector as −i+2j+5k.
Page 16 Problem 5 Answer
Given: The vector (2,4,0)
To write: The given vector by using the standard basis vectors for R2 and R3.
Evaluate to get the required solution.
Standard notation in R2
a=(a1,a2)=a1i+a2j
In R3: a=(a1,a2,a3)=a1i+a2j+a3k
Given vector in standard notation is: (2,4,0)
=2i+4j+0k
=2i+4j
Therefore, we can write the given vector by using the standard basis vectors as 2i+4j.
Page 16 Problem 6 Answer
Given: The vector i+j−3k
To write: The given vector without using the standard basis notation.
Evaluate to get the required answer.
Standard notation in R2:
a=(a1,a2)=a1i+a2j
In R3 : a=(a1,a2,a3)=a1i+a2j+a3k
Given the vector without standard notation is:
i+j−3k=(1,1,−3)
Therefore, we can write the given vector without using the standard basis notation as(1,1,−3).
Page 16 Problem 7 Answer
Given: The vector 9i−2j+√2k
To write: The given vector without using the standard basis notation.
Evaluate to get the final solution.
Standard notation in R2:
a=(a1,a2)=a1i+a2j
In R3: a=(a1,a2,a3)=a1i+a2j+a3k
Given vector without standard notation is:
9i−2j+√2k=(9,−2,√2)
Therefore, we can write the given vector without using the standard basis notation as(9,−2,√2).
Page 16 Problem 8 Answer
Given : The vector−3(2i−7k)
To find: The given vector without using the standard basis notation. Evaluate to get the final answer.
The Standard notation in R2:
a=(a1,a2)=a1i+a2j
In R3 :a=(a1,a2,a3)=a1i+a2j+a3k
Given vector without standard notation is:
−3(2i−7k)
=−6i+0j+21k
=(−6,0,21)
Therefore, we can write the given vector without using the standard basis notation as (−6,0,21).
Page 16 Problem 9 Answer
Given: The vector π i−j
To write: The given vector without using the standard basis notation.
Evaluate to get the final answer.
Standard notation in R2:
a=(a1,a2)=a1i+a2j
In R3 : a=(a1,a2,a3)=a1i+a2j+a3k
Given vector without standard notation is:
πi−j=(π,−1)
Therefore, we can write the given vector without using the standard basis notation as(π,−1).
Page 16 Problem 10 Answer
Given: πi−j in R3
To write: Without using the standard basis notation.
Evaluate to get the answer.
The standard notation in R2 is a=(a1,a2)
= a1i+a2j In R3:
a = (a1,a2,a3)=a1i+a2j+a3k
Given vector without standard notation is:
πi−j=πi−j+0k
=(π,−1,0)
Therefore, the given vector without standard notation is (π,−1,0)
Page 16 Problem 11 Answer
Given: a1=(1,1) and a2=(1,−1).
To write : c1a1+c2a2=b for b=(3,1).
Evaluate to get the solution.
First solve system c1a1+c2a2=b.
Substitute the values for a1 and a2.
c1(1,1)+c2(1,−1)=(3,1)
We obtain system:
c1+cv2=3
c1−c2=1
On solving these equations we get the value of c1 and c2: c1=2c2=1
Thus, b=2a1+a2.
Therefore, we can write b as b=2a1+a12.
Page 16 Problem 12 Answer
Given: a1=(1,1) and a2=(1,−1)
To write: Vector b=(3,−5) using part a of the question.
Evaluate to get the answer.
Solve system c{1}a{1}+c{2}a{2}=b.
c1(1,1)+c2(1,−1)=(3,−5)
We obtain system:
c1+c2=3
c1−c2=−5
Solution:
c1=−1 and c2=4
Thus, b=−a1+4a2
Therefore, we rewrite vector b as b=−a1+4a2.
Page 16 Problem 13 Answer
Given: a1=(1,1) and a2=(1,−1)
To show: Vector b=(b1,b2) in R2 in the form c1a1+c2a2
Evaluate to get the answer.
Solving the system c1a1+c2a2=b for an arbitrary vector b=(x,y).
c1(1,1)+c2(1,−1)=(x,y)
We obtain the system:
c1+c2=xc1−c2=y
Solution is
c1=x+y/2
c2=x−y/2
Thus, b can be written as a linear combination of a1 and a2.
Therefore, we have shown that b can be written as a linear combination of a1 and a2.
Page 16 Problem 14 Answer
Given: Vector i+3j+6k is parallel to the point (2,−1,5) and the line in R3 passes through the point.
To find: Give a set of parametric equations for the lines.
Evaluate to get the answer.
By using proposition, we can write:
r(y)=b+ta
=(2,−1,5)+t(1,3,−6)
The parametric equation is
x(t)=2+t
y(t)=−1+3t
z(t)=5−6t
Hence, the parametric equations for the given line are
x(t)=2+t
y(t)=−1+3t
z(t)=5−6t
Page 16 Problem 15 Answer
Given: Vector 5i−12j+k and point (12,−2,0)
To find: Give a set of parametric equations for the described lines.
Evaluate to get the answer.
Let us use a Proposition to write:
r(y)=b+ta=(12,−2,0)+t(5,−12,1)
The parametric equation is:
x(t)=12+5t
y(t)=−2−12t
z(t)=t
Hence, the set of parametric equations for the lines so described are x(t)=12+5t
y(t)=−2−12t
z(t)=t
Page 16 Problem 16 Answer
Given: Vector i−7j and points (2,−1)
To find: Give a set of parametric equations
Evaluate to get the answer.
Using Proposition, we can write:
r(y)=b+ta
=(2,−1)+t(1,−7)
Parametric equation is:
x(t)=2+t
y(t)=−1−7t
Therefore, the required set of parametric equations for the lines so described is x(t)=2+t
y(t)=−1−7t