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: 62


The magnitude is 8.25 and the direction angle is $284.0{}^\circ $.

Work Step by Step

If $\mathbf{v=}a\mathbf{i}-b\mathbf{j}$ is a vector then its magnitude is $\left\| \mathbf{v} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$. Use this definition of norm to get, $\begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -8 \right)}^{2}}} \\ & =\sqrt{4+64} \\ & =\sqrt{68} \\ & =8.25 \end{align}$ Hence, the magnitude of the vector $v$ is $8.25$. If $\mathbf{v=}a\mathbf{i}-b\mathbf{j}$ is a vector then its direction angle $\theta $ is $\theta ={{\tan }^{-1}}\left( \frac{b}{a} \right)$. So, $\begin{align} & \theta ={{\tan }^{-1}}\left( \frac{-8}{2} \right) \\ & ={{\tan }^{-1}}\left( -4 \right) \\ & =284.0{}^\circ \end{align}$. Hence, the angle is $284.0{}^\circ $.
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