## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 794: 68

#### Answer

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.

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