Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

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


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 $.
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