Sample Solution from
Multivariable Calculus
11th Edition
ISBN: 9781337275378
Chapter 11
Problem 1RE
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Textbook Problem

Writing Vectors in Different Forms In Exercises 1 and 2, let u = P Q ⇀ and v = P R ⇀ and (a) write u and v in component form, (b) write u and v as the linear combination of the standard unit vectors i and j, (c) find the magnitudes of u and v, and (d) find − 3 u + v .

P = ( 1 , 2 ) , Q = ( 4 , 1 ) , R = ( 5 , 4 )

Expert Solution

(a)

To determine

To calculate: For given P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→ Find acomponent form of u and v.

Answer

Solution:

For given P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→

Find a component form of u and vis 〈3,−1〉 and 〈4,2〉 respectively.

Explanation of Solution

Given:

P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→

Formula used:

If P(p1,p2) and Q(q1,q2) are the initial and terminal points of a directed line segment, then the component form of the vector u represented by PQ→ is

〈u1,u2〉=〈q1−p1,q2−p2〉

Calculation:

If

u2=q2−p2=1−2=−1

And

u =PQ→ and P=(1,2),Q=(4,1), then:

u1=q1−p1=4−1=3

The component form of u is 〈3,−1〉.

If v =PR→ and P=(1,2),R=(5,4), then:

v2=r2−p2=4−2=2

And

v1=r1−p1=5−1=4

The component form of v is 〈4,2〉.

Expert Solution

(b)

To determine

To calculate: For given P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→, find u and vas the linear combination of the standard unit vectors i and j.

Answer

Solution: For given P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→, u and vas the linear combination of the standard unit vectors i and j are the vectors u=3i−j and v=4i+2j

Explanation of Solution

Given:

P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→

Formula used:

If 〈v1,v2〉 is the component form of v, then the vector v=v1i+v2j is called a linear combination of i and j.

Calculation:

The vectors u=3i−j and v=4i+2j are the linear combinations of i and j.

As according to the calculation of part (a), a component form of u and vis 〈3,−1〉 and 〈4,2〉.

Expert Solution

(c)

To determine

To calculate: Forgiven P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→. Find magnitudes of u and v.

Answer

Solution:

Forgiven P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→

Magnitudes of u and vu is ‖u‖=10 and magnitude of v is ‖v‖=25.

Explanation of Solution

Given:

P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→

Formula used:

According to the Distance Formula, the length (or magnitude) of vectoru is:

‖u‖=(q1−p1)2+(q2−p2)2=u12+u22

Calculation:

As per part (a),

v=〈4,2〉. The length of vector v is:

‖v‖=v12+v22=42+22=16+4=20

u=〈3,−1〉. The length of vector u is:

‖u‖=u12+u22=32+(−1)2=9+1=10

and

Therefore, the magnitude of u is ‖u‖=10 and magnitude of v is ‖v‖=25.

Expert Solution

(d)

To determine

To calculate: For given P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→. Find the value of -3u+v.

Answer

Solution:

For given P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→. The value of -3u+v is 〈−5,5〉.

Explanation of Solution

Given:

P=(1,2),Q=(4,1),R=(5,4), u=PQ→,v=PR→

Formula used:

The vector sum of u and v is the vector:

u+v=〈u1+v1,u2+v2〉

The scalar multiple of c and u is the vector:

cu=〈cu1,cu2〉

Calculation:

Value of -3u+v using scalar multiplication and vector sum formulae:

As according to the calculation of part (a), a component form of u and v is 〈3,−1〉 and 〈4,2〉.

-3u+v=−3〈3,−1〉+〈4,2〉=〈−9,3〉+〈4,2〉=〈−5,5〉

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