Chapter 10: Q 31. (page 812)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information.

The given vectors are:

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

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

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

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

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=1,-5,-1,v=0,1,0

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

Probability and Statistics

Geometry

Theoretical and Mathematical Physics

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Applied Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.

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