Chapter 6, Problem 14RE

Solution

a)

To find: the direction angles of u and v.

The direction angle of $u=〈-2,4〉$ is $116.51°$ or $2.03$ radians and the direction angle of $v=〈6,4〉$ is $33.69°$ or $0.59$ radians.

Given information:

Given two vectors $u=〈-2,4〉,$

  $v=〈6,4〉$ .

Formula used:

• If $\theta$ is the direction angle of $v=〈x,y〉,$ then $x=\left|v\right|\cos \theta ,$ $y=\left|v\right|\cos \theta$ .
• The length of the vector $v=〈x,y〉$ is $\left|v\right|=\sqrt{x^2+y^2}$ .

Calculation:

The length of the vector $u=〈-2,4〉$:

Substituting $-2$ for $x$ and $4$ for $y$ in $\sqrt{x^2+y^2},$ the length of the vector $u$ is $\begin{array}{c}\sqrt{{\left(-2\right)}^2+{\left(4\right)}^2}=\sqrt{4+16}\\ =\sqrt{20}\\ =2\sqrt{5}\end{array}$

Thus, the length of the vector $u$ is $\left|u\right|=2\sqrt{5}$ .

The length of the vector $v=〈6,4〉:$

Substituting $6$ for $x$ and $4$ for $y$ in $\sqrt{x^2+y^2},$ the length of the vector $v$ is $\begin{array}{c}\sqrt{{\left(6\right)}^2+{\left(4\right)}^2}=\sqrt{36+16}\\ =\sqrt{52}\\ =2\sqrt{13}\end{array}$

Thus, the length of the vector $v$ is $\left|v\right|=2\sqrt{13}$ .

The direction angle of $u=〈-2,4〉:$

Substituting $-2$ for $x$ and $2\sqrt{5}$ for $\left|u\right|$ in $x=\left|u\right|\cos \theta ,$ then

  $\begin{array}{c}\cos \theta =\frac{-2}{2\sqrt{5}}\\ \theta ={\cos }^{-1}\left(-\frac{1}{\sqrt{5}}\right)\\ \approx 116.51°\end{array}$

The direction angle of $u=〈4,3〉$ is $116.51°$ or $2.03$ radians.

The direction angle of $v=〈6,4〉:$

If $\theta$ is the direction angle of $v,$ then $6=\left|v\right|\cos \theta$ . Thus, $\theta ={\cos }^{-1}\left(\frac{6}{\left|v\right|}\right)$ .

By substituting $2\sqrt{13}$ for $\left|v\right|$ , the required angle computed as follows:

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{6}{2\sqrt{13}}\right)\\ \approx 33.69°\end{array}$

The direction angle of $v=〈6,4〉$ is $33.69°$ or $0.59$ radians.

  

b)

To find: the angle between the vectors $u$ and $v.$

The angle between the vectors $u=〈-2,4〉$ and $v=〈6,4〉$ is $82.88°$ or $1.45$ radians.

Given information:

Given two vectors $u=〈-2,4〉,v=〈6,4〉.$

From part (a), the length of the vectors $u$ and $v$ respectively are $2\sqrt{5}$ and $2\sqrt{13}.$

Formula used:

• If $\theta$ is the angle between the nonzero vectors $u$ and $v$ , then $\cos \theta =\frac{u.v}{\left|u\right|\left|v\right|}$ .
• The dot product of two vectors $u=〈x_1,y_1〉,v=〈x_2,y_2〉$ is defined as $u.v=x_1{x}_2+y_1{y}_2.$

Calculation:

The dot product of $u$and $v$ :

The dot product of two vectors $u=〈-2,4〉,v=〈6,4〉$ is computed as follows:

  $\begin{array}{c}u.v=\left(-2\right)\left(6\right)+\left(4\right)\left(4\right)\\ =-12+16\\ =4.\end{array}$

Angle between the vectors:

If $\theta$ is the angle between the vectors $u$ and $v$ , then by substituting $u.v=4,\left|u\right|=2\sqrt{5}$ and $\left|v\right|=2\sqrt{13}$ , the value of $\cos \theta$ is given below:

  $\begin{array}{c}\cos \theta =\frac{u.v}{\left|u\right|\left|v\right|}\\ =\frac{4}{\left(2\sqrt{5}\right)\left(2\sqrt{13}\right)}\\ =\frac{1}{\sqrt{65}}.\end{array}$

Thus, the required angle is

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{65}}\right)\\ \approx 82.88°\end{array}$

Therefore, the angle between the vectors $u=〈-2,4〉$ and $v=〈6,4〉$ is $82.88°$ or $1.45$ radians.