Q 31.

Question

In Exercises 28-31, find ${proj}_{u} v and {proj}_{v} u$ .

$u = <1, - 5, - 1>, v = <0, 1, 0>$

Step-by-Step Solution

Verified

Answer

The values are ${proj}_{u} v = \frac{- 5}{27} and {proj}_{v} u = <0, - 5, 0>$ .

1Step 1. Given information.

The given vectors are:

$u = <1, - 5, - 1> and v = <0, 1, 0>$

2Step 2. Find the vector projection of v onto u .

${proj}_{u} v = \frac{u \cdot v}{{||u||}^{2}} u u = <1, - 5, - 1> and v = <0, 1, 0> u \cdot v = 1 (0) + (- 5) 1 + (- 1) 0 = 0 - 5 + 0 = - 5 ||u|| = \sqrt{1^{2} + {(- 5)}^{2} + {(- 1)}^{2}} = \sqrt{1 + 25 + 1} = \sqrt{27}$

Then,

${proj}_{u} v = \frac{u \cdot v}{{||u||}^{2}} u = \frac{- 5}{{(\sqrt{27})}^{2}} <1, - 5, - 1> = \frac{- 5}{27} <1, - 5, - 1>$

Therefore, ${proj}_{u} v = \frac{- 5}{27} <1, - 5, - 1>$ .

3Step 3. Find the vector projection of u onto v.

${proj}_{v} u = \frac{u \cdot v}{{||v||}^{2}} v u = <1, - 5, - 1> and v = <0, 1, 0> ||v|| = \sqrt{0^{2} + 1^{2} + 0^{2}} = \sqrt{1} = 1 {proj}_{v} u = \frac{u \cdot v}{{||v||}^{2}} v = \frac{- 5}{{(1)}^{2}} <0, 1, 0> = - 5 <0, 1, 0> = <0, - 5, 0>$

Therefore, ${proj}_{v} u = <0, - 5, 0>$ .

Q 30.

Q .32.

Other exercises in this chapter

Q 29.

In Exercises 28-31, find projuv and projvu.u=3,-4, v=16,12

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Q 30.

In Exercises 28-31, find projuv and projvu.u=3,0,1, v=2,2,-5

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Q .32.

In Exercises 32–36, (a) compute u.v(b) find the angle between u and v, and (c) find projuv

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Q .33.

In Exercises 32-36,(a) compute u . v(b) find the angle between u and v, and(c) find projuv

View solution