Chapter 10: Q 30. (page 812)

In Exercises 28-31, find ${proj}_{u} v and {proj}_{v} u$ .
$u = <3, 0, 1>, v = <2, 2, - 5>$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information.

The given vectors are:

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

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

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

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In Exercises 28-31, find ${proj}_{u} v and {proj}_{v} u$ .
$u = <3, 0, 1>, v = <2, 2, - 5>$

Short Answer

Step by step solution

Step 1. Given information.

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

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

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

Short Answer

Step by step solution

Step 1. Given information.

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

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

In Exercises 28-31, find ${proj}_{u} v and {proj}_{v} u$ .
$u = <3, 0, 1>, v = <2, 2, - 5>$