Q 26.

Question

In Exercises 24-27, find ${comp}_{u} v, {proj}_{u} v$ and the component of v orthogonal to u.

$u = <2, 0, - 5>, v = <- 3, 7, - 1>$

Step-by-Step Solution

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Answer

The values are ${comp}_{u} v = \frac{- 1}{\sqrt{29}}, {proj}_{u} v = \frac{- 1}{29} <2, 0, - 5>$ and the component of v orthogonal to u is $<\frac{- 85}{29}, 7, \frac{- 3}{29}>$ .

1Step 1. Given information.

The given vectors are:

$u = <2, 0, - 5> and v = <- 3, 7, - 1>$

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

${comp}_{u} v = \frac{u \cdot v}{||u||} u = <2, 0, - 5> and v = <- 3, 7, - 1> u \cdot v = 2 (- 3) + 0 (7) + (- 5) (- 1) = - 6 + 0 + 5 = - 1 ||u|| = \sqrt{2^{2} + 0^{2} + {(- 5)}^{2}} = \sqrt{4 + 0 + 25} = \sqrt{29}$

Then,

${comp}_{u} v = \frac{u \cdot v}{||u||} = \frac{- 1}{\sqrt{29}}$

Therefore, ${comp}_{u} v = \frac{- 1}{\sqrt{29}}$ .

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

${proj}_{u} v = \frac{u \cdot v}{{||u||}^{2}} u = \frac{- 1}{{(\sqrt{29})}^{2}} <2, 0, - 5> = \frac{- 1}{29} <2, 0, - 5>$

Therefore, the vector projection of v onto u is $\frac{- 1}{29} <2, 0, - 5>$ .

4Step 4. Find the component of v orthogonal to u .

The component of v orthogonal to u is $v - {proj}_{u} v$ .

$v - {proj}_{u} v = <- 3, 7, - 1> - (\frac{- 1}{29}) <2, 0, - 5> = <- 3, 7, - 1> + \frac{1}{29} <2, 0, - 5> = <- 3 + \frac{2}{29}, 7 + 0, - 1, - \frac{5}{29}> = <\frac{- 85}{29}, 7, \frac{- 3}{29}>$

Therefore, the component of v orthogonal to u is $<\frac{- 85}{29}, 7, \frac{- 3}{29}>$ .

Q 25.

Q 27.

Other exercises in this chapter

Q 24.

In Exercises 24-27, find compu v, proju v, and the component of v orthogonal to u.u=1,2, v=3,5

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Q 25.

In Exercises 24-27, find compuv , projuv and the component of v orthogonal to u.u=3,-1,2, v=-4,-6,3

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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 28.

In Exercises 28-31, find projuv and projvu.u=1,4, v=2,-3

View solution