Precalculus with Limits: A Graphing Approach
Precalculus with Limits: A Graphing Approach
7th Edition
ISBN: 9781305071711
Author: Ron Larson
Publisher: Brooks/Cole
Question
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Chapter 10, Problem 21RE

(a)

To determine

To find: The component form of the vector v with initial point (4,−2,1) and terminal point (−1,5,0) .

(a)

Expert Solution
Check Mark

Answer to Problem 21RE

The component form of the vector v with initial point (4,−2,1) and terminal point (−1,5,0) is 〈−5,7,−1〉 .

Explanation of Solution

Given information:

The vector v has initial point (4,−2,1) and terminal point (−1,5,0) .

Formula Used:

The component form of a vector v with initial point P(p1,p2,p3) and terminal point Q(q1,q2,q3) is v=〈q1−p1,q2−p2,q3−p3〉 .

Calculation:

Substitute 4 , −2 , 1 for p1 , p2 , p3 and −1 , 5 , 0 for q1 , q2 , q3 in the formula.

  v=〈q1−p1,q2−p2,q3−p3〉=〈−1−4,5−(−2),0−1〉=〈−5,7,−1〉

Therefore, the component form of the vector v with initial point (4,−2,1) and terminal point (−1,5,0) is 〈−5,7,−1〉 .

(b)

To determine

To find: The magnitude of vector v with initial point (4,−2,1) and terminal point (−1,5,0) .

(b)

Expert Solution
Check Mark

Answer to Problem 21RE

The magnitude of vector v with initial point (4,−2,1) and terminal point (−1,5,0) is 53 .

Explanation of Solution

Given information:

The vector v has initial point (4,−2,1) and terminal point (−1,5,0) .

Formula used:

The magnitude of any vector v with component form v=〈v1,v2,v3〉 is ‖v‖=v12+v22+v32 .

Calculation:

As calculated in part (a), the component form of the vector v is 〈−5,7,−1〉 .

Substitute −5 , 7 , −1 for v1 , v2 , v3 respectively in the formula.

  ‖v‖=v12+v22+v32=(−5)2+(7)2+(−1)2=25+49+1=75=53

Therefore, the magnitude of vector v with initial point (4,−2,1) and terminal point (−1,5,0) is 53 .

(c)

To determine

To find: A unit vector in the direction of v .

(c)

Expert Solution
Check Mark

Answer to Problem 21RE

The unit vector in the direction of v is 〈−33,7315,−315〉 .

Explanation of Solution

Given information:

The vector v has initial point (4,−2,1) and terminal point (−1,5,0) .

Formula used:

The unit vector in the direction of vector v is u=v‖v‖ .

Calculation:

As calculated in part (a), the component form of the vector v is 〈−5,7,−1〉 .

As calculated in part (b), the magnitude of the vector v is 53 .

Substitute 〈−5,7,−1〉 for v and 53 for ‖v‖ in the formula.

  u=v‖v‖=〈−5,7,−1〉53=〈−33,7315,−315〉

Therefore, the unit vector in the direction of v is 〈−33,7315,−315〉 .

Previous
Chapter 10, Problem 20RE
Next
Chapter 10, Problem 22RE

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

Ch. 10.1 - Prob. 1ECh. 10.1 - Prob. 2ECh. 10.1 - Prob. 3ECh. 10.1 - Prob. 4ECh. 10.1 - Prob. 5ECh. 10.1 - Prob. 6ECh. 10.1 - Prob. 7ECh. 10.1 - Prob. 8ECh. 10.1 - Prob. 9ECh. 10.1 - Prob. 10E
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