Chapter 11.2, Problem 8E

To determine

To Find:The magnitude of the vector and direction angle.

Answer to Problem 8E

The required magnitude of the vector is $\left|v\right|=4$ and angle $\theta =-{30}^{\circ }$ .

Explanation of Solution

Given Information:

Vector is $\left(-2,-2\sqrt{3}\right)$

Formula Used:

Magnitude formula

  $\left|v\right|=\sqrt{a^2+b^2}$  ........ (1)

Here, $a$ and $b$ are two points.

  $\cos \theta =\frac{a}{\left|v\right|}$ , $\sin \theta =\frac{b}{\left|v\right|}$  ........ (2)

Calculation:

Substitute $-2,-2\sqrt{3}$ for $a,b$ in equation (1)

  $\begin{array}{c}\left|v\right|=\sqrt{{\left(-2\right)}^2+{\left(-2\sqrt{3}\right)}^2}\\ =4\end{array}$

Substitute $-2,-2\sqrt{3}$ for $a,b$ and $4$ for $\left|v\right|$ in equation (2)

  $\begin{array}{l}\cos \theta =\frac{-2}{4}=\sin \theta =\frac{-2\sqrt{3}}{4}\\ \theta ={\cos }^{-1}-\frac{1}{2}={\sin }^{-1}-\frac{\sqrt{3}}{2}\\ \theta =-{30}^{\circ }\end{array}$

Conclusion:

Hence, the required magnitude of the vector is $\left|v\right|=4$ and angle $\theta =-{30}^{\circ }$ .