Q .33. - Calculus

Question

In Exercises 32-36,

(a) compute u . v

(b) find the angle between u and v, and

(c) find ${proj}_{u}v$

Step-by-Step Solution

Answer

Part a)$u.v=-1$

Part b)The angle between the vectors is $\theta ={\cos }^{-1}\left(\frac{-1}{2\sqrt{2451}}\right)$

Part c)${\text{ Value of proj }}_{\mathrm{n}}\mathrm{v}\text{ is }\frac{-1}{38}(-2,3,5)\text{. }$

Step 1:Given information Part a)

u and v are vectors 

Step 2 Part a) Simplification

$\text{ Consider the vector }\mathbf{u}=⟨-2,3,5⟩,\mathbf{v}=⟨13,-5,8⟩$

$\text{ given by }\mathbf{u}\mathbf{.}\mathbf{v}=u_1{v}_1+u_2{v}_2+w_1{w}_2$

$\mathbf{u}·\mathbf{v}=-2\left(13\right)+3(-5)+5\left(8\right)$

$=-26-15+40$$=-1$

Step 3:Part b) Given information

$\text{ the vector }\mathbf{u}=⟨-2,3,5⟩,\mathbf{v}=⟨13,-5,8⟩$

Step 4:Part b)Simplification

$u.v=-1$

$\text{ Also, }\mathbf{u}=⟨-2,3,5⟩$

$\Vert \mathbf{u}\Vert =\sqrt{(-2{)}^2+3^2+5^2}$

$=\sqrt{38}$

$\text{ Also, }\mathbf{v}=⟨13,-5,8⟩$

$\Vert \mathbf{v}\Vert =\sqrt{{13}^2+(-5{)}^2+8^2}$

$\sqrt{258}$

now

$\cos \theta =\frac{-1}{\sqrt{38}\sqrt{358}}$

$=\frac{-1}{2\sqrt{2451}}$

Therefore

$\theta ={\cos }^{-1}\left(\frac{-1}{2\sqrt{2451}}\right)$

Hence, the angle between the vectors is $\theta ={\cos }^{-1}\left(\frac{-1}{2\sqrt{2451}}\right)$

Step 5 Part c):Given information

$\text{ the vector }\mathbf{u}=⟨-2,3,5⟩,\mathbf{v}=⟨13,-5,8⟩$

Step 6:Part c) Simplification

$u.v=-1$ (calculated)

$\Vert \mathbf{u}\Vert =\sqrt{38}$ (calculated)

$\text{ Let }\mathbf{u}\text{ be any non-zero vector, then the vector projection of }\mathbf{v}\text{ onto }\mathbf{u}\text{ is given by }$

${proj}_{\mathbf{u}}\mathbf{v}=\frac{\mathbf{u}·\mathbf{v}}{\Vert \mathbf{u}{\Vert }^2}\mathbf{u}\text{.Now, substitute the values in }{proj}_{\mathbf{u}}\mathbf{v}=\frac{\mathbf{u}·\mathbf{v}}{\Vert \mathbf{u}{\Vert }^2}\mathbf{u}$

${proj}_{\mathrm{u}}\mathbf{v}=\frac{-1}{(\sqrt{38}{)}^2}⟨-2,35⟩$

$=\frac{-1}{38}⟨-2,3,5⟩$

${\text{ Hence, value of proj }}_u\mathrm{v}\text{ is }\frac{-1}{38}(-2,3,5)\text{. }$