Chapter 6, Problem 13RE

Solution

a)

To find: the direction angles of u and v.

The direction angle of $u=〈4,3〉$ is $36.87°$ or $0.64$ radians, and the direction angle of $v=〈2,5〉$ is $68.2°$ or 1.19 radians.

Given information:

Given two vectors $u=〈4,3〉,$

  $v=〈2,5〉$ .

Formula used:

• If $\theta$ is the direction angle of $v=〈x,y〉,$ then $\tan \theta =\frac{y}{x}$ .

Calculation:

The direction angle of $u=〈4,3〉:$

Substituting $4$ for $x$ and $3$ for $y$ in $\tan \theta =\frac{y}{x}$ , then

  $\begin{array}{c}\tan \theta =\frac{3}{4}\\ \theta ={\tan }^{-1}\left(\frac{3}{4}\right)\\ \approx 36.87°\\ \end{array}$

The direction angle of $u=〈4,3〉$ is $36.87°$ or $0.64$ radians.

The direction angle of $v=〈2,5〉:$

Substituting $2$ for $x$ and $5$ for $y$ in $\tan \theta =\frac{y}{x}$ ,

  $\begin{array}{c}\tan \theta =\frac{5}{2}\\ \theta ={\tan }^{-1}\left(\frac{5}{2}\right)\\ \approx 68.2°\\ \end{array}$

The direction angle of $v=〈2,5〉$ is $68.2°$ or 1.19 radians.

  

Figure (1)

b)

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

The angle between the vectors $u=〈4,3〉$ and $v=〈2,5〉$ is $31.33°$ or $0.55$ radians.

Given information:

Given two vectors $u=〈4,3〉,$

  $v=〈2,5〉$ .

From part (a), the length of the vectors $u$ and $v$ respectively are $5$ and $\sqrt{29}.$

Formula used:

• The length of the vector $v=〈x,y〉$ is $\left|v\right|=\sqrt{x^2+y^2}.$
• 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 length of the vector $u=〈4,3〉$:

Substituting $4$ for $x$ and $3$ for $y$ in $\sqrt{x^2+y^2},$ the length of the vector $u$ is $\begin{array}{c}\sqrt{{\left(4\right)}^2+{\left(3\right)}^2}=\sqrt{16+9}\\ =\sqrt{25}\\ =5\end{array}$

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

The length of the vector $v=〈2,5〉$:

Substituting $2$ for $x$ and $5$ for $y$ in $\sqrt{x^2+y^2},$ the length of the vector $v$ is $\begin{array}{c}\sqrt{{\left(2\right)}^2+{\left(5\right)}^2}=\sqrt{4+25}\\ =\sqrt{29}\end{array}$

Thus, the length of the vector $v$ is $\left|v\right|=\sqrt{29}$ .

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

The dot product of two vectors $u=〈4,3〉,$

  $v=〈2,5〉$ is computed as follows:

  $\begin{array}{c}u.v=\left(4\right)\left(2\right)+\left(3\right)\left(5\right)\\ =8+15\\ =23\end{array}$

Angle between the vectors:

If $\theta$ is the angle between the vectors $u$ and $v$ , then by substituting $u.v=23,$

  $\left|u\right|=5$ and $\left|v\right|=\sqrt{29}$ , the value of $\cos \theta$ is given below:

  $\begin{array}{c}\cos \theta =\frac{u.v}{\left|u\right|\left|v\right|}\\ =\frac{23}{\left(5\right)\left(\sqrt{29}\right)}\\ =\frac{23}{5\sqrt{29}}.\end{array}$

Thus, the required angle is

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{23}{5\sqrt{29}}\right)\\ \approx 31.33°\\ \approx 0.55\end{array}$

Therefore, the angle between the vectors $u=〈4,3〉$ and $v=〈2,5〉$ is $31.33°$ or $0.55$ radians.