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: 17


${{79.70}^{{}^\circ }}$

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, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( 2\mathbf{i-j} \right)\cdot \left( 3\mathbf{i+}4\mathbf{j} \right)}{\left( \sqrt{{{2}^{2}}+{{1}^{2}}} \right)\left( \sqrt{{{3}^{2}}\mathbf{+}{{4}^{2}}} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{2\cdot 3+\left( -1 \right)\cdot 4}{\left( \sqrt{5} \right)\left( \sqrt{25} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{2}{5\sqrt{5}} \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{2}{5\sqrt{5}} \right) \\ & ={{79.70}^{{}^\circ }} \end{align}$ Hence, the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta ={{79.70}^{{}^\circ }}$.
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