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: 20


${{100.30}^{{}^\circ }}$

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, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( \mathbf{i}+2\mathbf{j} \right)\cdot \left( 4\mathbf{i}-3\mathbf{j} \right)}{\left( \sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}} \right)\left( \sqrt{{{4}^{2}}\mathbf{+}{{\left( -3 \right)}^{2}}} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{1\cdot 4+2\cdot \left( -3 \right)}{\left( \sqrt{5} \right)\left( \sqrt{25} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{-2}{\sqrt{125}} \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{-2}{\sqrt{125}} \right) \\ & ={{100.30}^{{}^\circ }} \end{align}$ Hence, the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta ={{100.30}^{{}^\circ }}$.
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