Precalculus (6th Edition) Blitzer

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

Chapter 6 - Test - Page 800: 18


The angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta \approx 138{}^\circ $.

Work Step by Step

We have $\begin{align} & \cos \theta =\frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\left\| \mathbf{w} \right\|} \\ & =\frac{\left( -5\mathbf{i}+2\mathbf{j} \right)\cdot \left( 2\mathbf{i}-4\mathbf{j} \right)}{\left( \sqrt{{{\left( -5 \right)}^{2}}+{{2}^{2}}} \right)\cdot \left( \sqrt{{{2}^{2}}+{{\left( -4 \right)}^{2}}} \right)} \\ & =\frac{\left( -5\left( 2 \right) \right)+2\left( -4 \right)}{\left( \sqrt{29} \right)\cdot \left( \sqrt{20} \right)} \\ & =-\frac{18}{\left( \sqrt{580} \right)} \end{align}$ So, $\begin{align} & \theta ={{\cos }^{-1}}\left( -\frac{18}{\left( \sqrt{580} \right)} \right) \\ & \approx 138{}^\circ \end{align}$ Therefore, the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta \approx 138{}^\circ $.
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