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 793: 48


The vectors $\mathbf{v}$ and $\mathbf{w}$ neither parallel nor orthogonal.

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, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left| \mathbf{v} \right|\left| \mathbf{w} \right|} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( -2\mathbf{i}+3\mathbf{j} \right)\cdot \left( -6\mathbf{i}-9\mathbf{j} \right)}{\left( \sqrt{{{\left( -2 \right)}^{2}}+{{3}^{2}}} \right)\left( \sqrt{{{\left( -6 \right)}^{2}}\mathbf{+}{{\left( -9 \right)}^{2}}} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{\left( -2 \right)\cdot \left( -6 \right)+3\cdot \left( -9 \right)}{\left( \sqrt{13} \right)\left( \sqrt{117} \right)} \right) \\ & ={{\cos }^{-1}}\left( \frac{12-27}{\sqrt{1521}} \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \theta ={{\cos }^{-1}}\left( \frac{12-27}{\sqrt{1521}} \right) \\ & ={{\cos }^{-1}}\left( \frac{-15}{39} \right) \\ & ={{\cos }^{-1}}\left( -0.384 \right) \\ & ={{113}^{{}^\circ }} \end{align}$ Since, the angle between $\mathbf{v}$ and $\mathbf{w}$ is ${{113}^{{}^\circ }}$, $\mathbf{v}$ and $\mathbf{w}$ are neither parallel nor orthogonal vectors. Hence, $\mathbf{v}$ and $\mathbf{w}$ are neither parallel nor orthogonal.
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