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: 22


${{38.70}^{{}^\circ }}$

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, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( 0\mathbf{i}+3\mathbf{j} \right)\cdot \left( 4\mathbf{i}+5\mathbf{j} \right)}{\left( \sqrt{{{\left( 0 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \right)\left( \sqrt{{{4}^{2}}\mathbf{+}{{\left( 5 \right)}^{2}}} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{0\cdot 4+3\cdot 5}{\left( \sqrt{9} \right)\left( \sqrt{41} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{15}{\sqrt{369}} \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{15}{\sqrt{369}} \right) \\ & ={{38.70}^{{}^\circ }} \end{align}$ Hence, the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta ={{38.70}^{{}^\circ }}$.
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