Chapter 11, Problem 1RE

### Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095

Chapter
Section

### Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
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 )

(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.

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

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=q1p1,q2p2

Calculation:

If

u2=q2p2=12=1

And

u =PQ and P=(1,2),Q=(4,1), then:

u1=q1p1=41=3

The component form of u is 3,1.

If v =PR and P=(1,2),R=(5,4), then:

v2=r2p2=42=2

And

v1=r1p1=51=4

The component form of v is 4,2.

(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.

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=3ij and v=4i+2j

Explanation

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=3ij 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.

(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.

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

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=(q1p1)2+(q2p2)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.

(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.

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

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=33,1+4,2=9,3+4,2=5,5

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