Q .32. - Calculus

Question

In Exercises 32–36, 

(a) compute $u.v$

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

 (c) find $pro{j}_{u}v$

Step-by-Step Solution

Answer

Part a)$u.v=37$

Part b) angle is $\theta ={\cos }^{-1}\left(\frac{37}{\sqrt{1378}}\right)$

Part c)$\text{ The value of }{proj}_{\mathrm{u}}\mathrm{v}\text{ is }\frac{37}{26}⟨1,5⟩\text{. }$

Step 1:Given information Part a)

u and v are vectors

Step 2 Part a) Simplification

$\text{ Consider the vector }\mathbf{u}=⟨1,5⟩,\mathbf{v}=⟨2,7⟩\text{. }$

$\mathbf{u}.\mathit{v}=u_1{v}_1+u_2{v}_2.$

$\mathbf{u}·\mathbf{v}=1\left(2\right)+5\left(7\right)$

$=37$

Step 3:Part b) Given information

$u,v$ are given vectors

Step 4:Part b)Simplification

$u.v=37$

$\mathrm{Also},\mathbf{u}=⟨1,5⟩$

Therefore

$\Vert \mathbf{u}\Vert =\sqrt{1^2+5^2}$

$=\sqrt{26}$

$\text{ Also, }\mathbf{v}=⟨2,7⟩$

$\Vert \mathbf{v}\Vert =\sqrt{2^2+7^2}$

$=\sqrt{53}$

now,

$\cos \theta =\frac{37}{\sqrt{26}\sqrt{53}}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{37}{\sqrt{1378}}\right)$

Hence, the angle between the vectors is ${\cos }^{-1}\left(\frac{37}{\sqrt{1378}}\right)$

Step 5 Part c):Given information

$\text{ the vector }\mathbf{u}=⟨1,5⟩,\mathbf{v}=⟨2,7⟩$

Step 6:Part c) Simplification

$\Rightarrow u.v=37$ calculated

and

$\Vert \mathbf{u}\Vert =\sqrt{26}$

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

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

${proj}_{\mathrm{u}}\mathbf{v}=\frac{37}{(\sqrt{26}{)}^2}⟨1,5⟩$

$=\frac{37}{26}⟨1,5⟩$

$\text{ Hence, the value of }{proj}_{\mathrm{u}}\mathrm{v}\text{ is }\frac{37}{26}⟨1,5⟩\text{. }$