Chapter 10: Q 28. (page 812)

In Exercises 28-31, find ${proj}_{u} v and {proj}_{v} u$ .
$u = <1, 4>, v = <2, - 3>$

Short Answer

Expert verified

The values are ${proj}_{u} v = \frac{- 10}{17} <1, 4> and {proj}_{v} u = \frac{- 10}{13} <2, - 3>$ .

Step by step solution

Step 1. Given information.

The given vectors are:

$u = <1, 4>, v = <2, - 3>$

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

${proj}_{u} v = \frac{u \cdot v}{{||u||}^{2}} u u = <1, 4>, v = <2, - 3> u \cdot v = 1 (2) + 4 (- 3) = 2 - 12 = - 10 ||u|| = \sqrt{1^{2} + 4^{2}} = \sqrt{1 + 16} = \sqrt{17}$

Then

${proj}_{u} v = \frac{u \cdot v}{{||u||}^{2}} u = \frac{- 10}{{(\sqrt{17})}^{2}} <1, 4> = \frac{- 10}{17} <1, 4>$

Therefore, ${proj}_{u} v = \frac{- 10}{17} <1, 4>$ .

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

${proj}_{v} u = \frac{u \cdot v}{{||v||}^{2}} v u = <1, 4>, v = <2, - 3> ||v|| = \sqrt{2^{2} + {(- 3)}^{2}} = \sqrt{4 + 9} = \sqrt{13}$

Then,

${proj}_{v} u = \frac{u \cdot v}{{||v||}^{2}} v = \frac{- 10}{{(\sqrt{13})}^{2}} <2, - 3> = \frac{- 10}{13} <2, - 3>$

Therefore, ${proj}_{v} u = \frac{u \cdot v}{{||v||}^{2}} v = \frac{- 10}{13} <2, - 3>$ .

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In Exercises 28-31, find ${proj}_{u} v and {proj}_{v} u$ .
$u = <1, 4>, v = <2, - 3>$

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=1,4,v=2,-3

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

Applied Mathematics

Discrete Mathematics

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Logic and Functions

Study anywhere. Anytime. Across all devices.

In Exercises 28-31, find ${proj}_{u} v and {proj}_{v} u$ .
$u = <1, 4>, v = <2, - 3>$