Q 28.

Question

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

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

Step-by-Step Solution

Verified

Answer

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

1Step 1. Given information.

The given vectors are:

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

2Step 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, ${proj}_{u} v = \frac{- 10}{17} <1, 4>$ .

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

Q 27.

Q 29.

Other exercises in this chapter

Q 26.

In Exercises 24-27, find compuv , projuv and the component of v orthogonal to u.u=2,0,-5, v=-3,7,-1

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Q 27.

In Exercises 24-27, find compuv , projuv and the component of v orthogonal to u. u=3,1,-2, v=-6,-2,

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Q 29.

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

View solution

Q 30.

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

View solution