Question

In Exercises 32–36,

(a) compute$u.v$

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

(c) find$pro{j}_{u}v$

Short 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-by-step solution

Step 1:Given information Part a)

u and v are vectors

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

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

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

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

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

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

$=\frac{37}{26}⟨1,5⟩$

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