Chapter 11.2, Problem 6E

To determine

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

Answer to Problem 6E

The required magnitude of the vector is $\left|v\right|=2$ and angle $\theta ={135}^{\circ }$ .

Explanation of Solution

Given Information:

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

Formula Used:

Magnitude formula

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

Here, $\left|v\right|$ is the magnitude, $a$ and $b$ are two points.

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

Calculation:

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

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

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

  $\begin{array}{l}\cos \theta =\frac{-\sqrt{2}}{2}=\sin \theta =\frac{\sqrt{2}}{2}\\ \theta ={\cos }^{-1}-\frac{1}{\sqrt{2}}={\sin }^{-1}\frac{1}{\sqrt{2}}\\ \theta ={135}^{\circ }\end{array}$

Conclusion:

Hence, the required magnitude of the vector is $\left|v\right|=2$ and angle $\theta ={135}^{\circ }$ .