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 794: 68


Two vectors are said to be parallel vectors if they have the same direction or are in exactly opposite directions, such that the angle between them is ${{0}^{\circ }}$.

Work Step by Step

Consider vectors $\mathbf{v}=2\mathbf{i}+\mathbf{j}+3\mathbf{k}$, $\mathbf{w}=\mathbf{i}+2\mathbf{j}+\mathbf{k}$, and $\mathbf{v}\cdot \mathbf{w}=9.165$. Find the magnitude of vector v as follows: $\begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\ & =\sqrt{4+1+9} \\ & =\sqrt{14} \end{align}$ $\left\| \mathbf{v} \right\|=\sqrt{14}$ Similarly, find the magnitude of vector w as follows: $\begin{align} & \left\| \mathbf{w} \right\|=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\ & =\sqrt{1+4+1} \\ & =\sqrt{6} \end{align}$ $\left\| \mathbf{w} \right\|=\sqrt{6}$ Now, find the cosine of the angle between their directions as follows: $\begin{align} & \cos \theta =\left( \frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\text{ }\left\| \mathbf{w} \right\|} \right) \\ & =\left( \frac{9.165}{\sqrt{14}\sqrt{6}} \right) \\ & =\left( \frac{9.165}{9.165} \right) \\ & =1 \end{align}$ So that $\begin{align} & \theta ={{\cos }^{-1}}\left( 1 \right) \\ & ={{0}^{\circ }} \end{align}$ Hence, the angle between two vectors is ${{0}^{\circ }}$.Therefore, it can be concluded that the vectors are parallel to each other.
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.