## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 782: 61

#### Answer

The magnitude is $18.03$ and the angle is $123.7{}^\circ$.

#### Work Step by Step

If $v=a\mathbf{i}+b\mathbf{j}$ is a vector, then its magnitude is $\left\| v \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$. Use this definition of norm to get \begin{align} & \left\| v \right\|=\sqrt{{{\left( -10 \right)}^{2}}+{{\left( 15 \right)}^{2}}} \\ & =\sqrt{100+225} \\ & =\sqrt{325} \\ & =18.03 \end{align} Hence, the magnitude of the vector $v$ is 18.03. If $v=a\mathbf{i}+b\mathbf{j}$ is a vector, then its direction angle $\theta$ is $\theta ={{\tan }^{-1}}\left( \frac{b}{a} \right)$. So, we have \begin{align} & \theta ={{\tan }^{-1}}\left( \frac{15}{-10} \right) \\ & =123.7{}^\circ \end{align}. Hence, the angle is $123.7{}^\circ$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.