Q 24.

Question

In Exercises 24-27, find comp_uv, proj_uv, and the component of v orthogonal to u.

$u = <1, 2>, v = <3, 5>$

Step-by-Step Solution

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Answer

The values are ${comp}_{u} v = \frac{13}{\sqrt{5}}, {proj}_{u} v = \frac{13}{5} <1, 2>$ and the component of v orthogonal to u is $<\frac{2}{5}, \frac{- 1}{5}>$ .

1Step 1. Given information.

The given vectors are:

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

2Step 2. Find comp u v.

${comp}_{u} v = \frac{u \cdot v}{||u||}$ is the projection of v onto u.

$u = <1, 2> and v = <3, 5> u \cdot v = 1 (3) + 2 (5) = 3 + 10 = 13 ||u|| = \sqrt{1^{2} + 2^{2}} = \sqrt{1 + 4} = \sqrt{5}$

then, ${comp}_{u} v = \frac{u \cdot v}{||u||}$ is:

$= \frac{13}{\sqrt{5}}$

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

${proj}_{u} v = \frac{u \cdot v}{{||u||}^{2}} u = \frac{13}{{(\sqrt{5})}^{2}} <1, 2> = \frac{13}{5} <1, 2>$

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

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

$v - {proj}_{u} v = <3, 5> - \frac{13}{5} <1, 2> = <3, 5> - <\frac{13}{5}, \frac{26}{5}> = <\frac{2}{5}, \frac{- 1}{5}>$

Therefore, the component of v orthogonal to u is $<\frac{2}{5}, \frac{- 1}{5}>$ .

Q 23.

Q 25.

Other exercises in this chapter

Q 22.

In Exercises 20-23, find the dot product of the given pairs of vectors and the angle between the two vectors.u=3,-1,2, v=-4,-6,3

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Q 23.

In Exercises 20-23, find the dot product of the given pairs of vectors and the angle between the two vectors.u=-5,1,3, v=-3,2,7

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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 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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