Q 30.

Question

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

$u = <3, 0, 1>, v = <2, 2, - 5>$

Step-by-Step Solution

Verified

Answer

The values are ${proj}_{u} v = \frac{1}{10} <3, 0, 1> and {proj}_{v} u = \frac{1}{33} <2, 2, - 5>$ .

1Step 1. Given information.

The given vectors are:

$u = <3, 0, 1> and v = <2, 2, - 5>$

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

${proj}_{u} v = \frac{u \cdot v}{{||u||}^{2}} u u = <3, 0, 1> and v = <2, 2, - 5> u \cdot v = 3 (2) + 0 (2) + 1 (- 5) = 6 + 0 - 5 = 1 ||u|| = \sqrt{3^{2} + 0^{2} + 1^{2}} = \sqrt{9 + 0 + 1} = \sqrt{10}$

Then,

${proj}_{u} v = \frac{u \cdot v}{{||u||}^{2}} u = \frac{1}{{(\sqrt{10})}^{2}} <3, 0, 1> = \frac{1}{10} <3, 0, 1>$

Therefore, ${proj}_{u} v = \frac{1}{10} <3, 0, 1>$ .

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

${proj}_{v} u = \frac{u \cdot v}{{||v||}^{2}} v u = <3, 0, 1> and v = <2, 2, - 5> ||v|| = \sqrt{2^{2} + 2^{2} + {(- 5)}^{2}} = \sqrt{4 + 4 + 25} = \sqrt{33}$

Then,

${proj}_{v} u = \frac{u \cdot v}{{||v||}^{2}} v = \frac{1}{{(\sqrt{33})}^{2}} <2, 2, - 5> = \frac{1}{33} <2, 2, - 5>$

Therefore, ${proj}_{v} u = \frac{1}{33} <2, 2, - 5>$ .

Q 29.

Q 31.

Other exercises in this chapter

Q 28.

In Exercises 28-31, find projuv and projvu.u=1,4, v=2,-3

View solution

Q 29.

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

View solution

Q 31.

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

View solution

Q .32.

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

View solution