Precalculus (6th Edition) Blitzer

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

Chapter 6 - Review Exercises - Page 800: 97


$\mathbf{v}\cdot \mathbf{w}=2\text{ and }\theta \approx 86.1{}^\circ $.

Work Step by Step

Dot product of vectors: For the two given vectors $\mathbf{v}={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}$ and $\mathbf{w}={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}$ $\mathbf{v}\cdot \mathbf{w}={{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}$ and $\cos \theta =\frac{\left\| \mathbf{v} \right\|\cdot \left\| \mathbf{w} \right\|}{\mathbf{v}\cdot \mathbf{w}}$ where $\theta $ is the angle between the vectors $\mathbf{v}$ and $\mathbf{w}$. Here, ${{a}_{1}}=2,{{a}_{2}}=7,{{b}_{1}}=3,{{b}_{2}}=-4$ $\begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{2}^{2}}+{{3}^{2}}} \\ & =\sqrt{4+9} \\ & =\sqrt{13} \end{align}$ $\begin{align} & \left\| \mathbf{w} \right\|=\sqrt{{{7}^{2}}+{{\left( -4 \right)}^{2}}} \\ & =\sqrt{49+16} \\ & =\sqrt{65} \end{align}$ So, $\begin{align} & \mathbf{v}\cdot \mathbf{w}=2\left( 7 \right)+3\left( -4 \right) \\ & =14-12 \\ & =2 \end{align}$ And also, $\begin{align} & \cos \theta =\frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\cdot \left\| \mathbf{w} \right\|} \\ & =\frac{2}{\sqrt{13}\cdot \sqrt{65}} \\ & =\frac{2}{\sqrt{845}} \end{align}$ Using a calculator we get $\begin{align} & \cos \theta =\frac{2}{\sqrt{845}} \\ & \theta ={{\cos }^{-1}}\left( \frac{2}{\sqrt{845}} \right) \\ & \theta \approx 86.1{}^\circ \end{align}$ Hence, $\mathbf{v}\cdot \mathbf{w}=2$ and the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta \approx 86.1{}^\circ $.
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