Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 794: 67


The angle between two vectors $\theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\text{ }\left\| \mathbf{w} \right\|} \right)$.

Work Step by Step

Take $\mathbf{v}=2\mathbf{i}+\mathbf{j}+3\mathbf{k}$, $\mathbf{w}=\mathbf{i}+2\mathbf{j}+\mathbf{k}$, and $\mathbf{v}\cdot \mathbf{w}=9.165$. Now, find the magnitude of vector v as follows: $\begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\ & =\sqrt{4+1+9} \\ & =\sqrt{14} \end{align}$ Therefore, $\left\| \mathbf{v} \right\|=\sqrt{14}$. Similarly, find the magnitude of the vector $w$ as follows: $\begin{align} & \left\| \mathbf{w} \right\|=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\ & =\sqrt{1+4+1} \\ & =\sqrt{6} \end{align}$ Therefore, $\left\| \mathbf{w} \right\|=\sqrt{6}$. Next, find the cosine of the angle between their directions as follows: $\begin{align} & \cos \theta =\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\text{ }\left\| \mathbf{w} \right\|} \right) \\ & =\left( \frac{9.165}{\sqrt{14}\sqrt{6}} \right) \\ & =\left( \frac{9.165}{9.165} \right) \\ & =1 \end{align}$ So, $\cos \theta =1$. Therefore, $\begin{align} & \theta ={{\cos }^{-1}}\left( 1 \right) \\ & ={{0}^{\circ }} \end{align}$ Hence, the angle between two vectors is ${{0}^{\circ }}$.
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