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

Theoretical and Mathematical Physics

Discrete Mathematics

Geometry

Pure Maths

Decision Maths

Mechanics Maths

Study anywhere. Anytime. Across all devices.

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