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 792: 18


${{48.40}^{{}^\circ }}$

Work Step by Step

Let the angle between $\mathbf{v}$ and $\mathbf{w}$ be $\theta $ such that the angle between the vectors $\mathbf{v}$ and $\mathbf{w}$ can be obtained using the formula $\theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right)$ as, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( -2\mathbf{i}+5\mathbf{j} \right)\cdot \left( 3\mathbf{i+}6\mathbf{j} \right)}{\left( \sqrt{{{\left( -2 \right)}^{2}}+{{\left( 5 \right)}^{2}}} \right)\left( \sqrt{{{3}^{2}}\mathbf{+}{{6}^{2}}} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( -2 \right)\cdot 3+5\cdot 6}{\left( \sqrt{29} \right)\left( \sqrt{45} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{24}{\sqrt{1305}} \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{24}{\sqrt{1305}} \right) \\ & ={{48.40}^{{}^\circ }} \end{align}$ Hence, the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta ={{48.40}^{{}^\circ }}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.