## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Review Exercises - Page 800: 99

#### Answer

$\mathbf{v}\cdot \mathbf{w}=1\text{ }and\text{ }\theta \approx 71.6{}^\circ$.

#### Work Step by Step

Dot product of vectors: For two given vectors $\mathbf{v}={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}$ and $\mathbf{w}={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}$ $\mathbf{v}\cdot \mathbf{w}={{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}$ and $\cos \theta =\frac{\left\| \mathbf{v} \right\|\cdot \left\| \mathbf{w} \right\|}{\mathbf{v}\cdot \mathbf{w}}$ where $\theta$ is the angle between the vectors $\mathbf{v}$ and $\mathbf{w}$. Here, ${{a}_{1}}=2,{{a}_{2}}=1,{{b}_{1}}=1,{{b}_{2}}=-1$ \begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{2}^{2}}+{{1}^{2}}} \\ & =\sqrt{4+1} \\ & =\sqrt{5} \end{align} \begin{align} & \left\| \mathbf{w} \right\|=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}} \\ & =\sqrt{1+1} \\ & =\sqrt{2} \end{align} So, \begin{align} & \mathbf{v}\cdot \mathbf{w}=2\left( 1 \right)+1\left( -1 \right) \\ & =2-1 \\ & =1 \end{align} And also, \begin{align} & \cos \theta =\frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\cdot \left\| \mathbf{w} \right\|} \\ & =\frac{1}{\sqrt{5}\cdot \sqrt{2}} \\ & =\frac{1}{\sqrt{10}} \end{align} Use a calculator to get \begin{align} & \cos \theta =\frac{1}{\sqrt{10}} \\ & \theta ={{\cos }^{-1}}\left( \frac{1}{\sqrt{10}} \right) \\ & \theta \approx 71.6{}^\circ \end{align} Hence, $\mathbf{v}\cdot \mathbf{w}=1$ and the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta \approx 71.6{}^\circ$.

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