# Chapter 6 - Review Exercises - Page 800: 98

$\mathbf{v}\cdot \mathbf{w}=-32\text{ and }\theta \approx 124.8{}^\circ$.

#### 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, ${{a}_{1}}=2,{{a}_{2}}=6,{{b}_{1}}=4,{{b}_{2}}=-11$ \begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{2}^{2}}+{{4}^{2}}} \\ & =\sqrt{4+16} \\ & =\sqrt{20} \end{align} \begin{align} & \left\| \mathbf{w} \right\|=\sqrt{{{6}^{2}}+{{\left( -11 \right)}^{2}}} \\ & =\sqrt{36+121} \\ & =\sqrt{157} \end{align} So, \begin{align} & \mathbf{v}\cdot \mathbf{w}=2\left( 6 \right)+4\left( -11 \right) \\ & =12-44 \\ & =-32 \end{align} And also, \begin{align} & \cos \theta =\frac{\mathbf{v}\cdot \mathbf{w}}{\left\| \mathbf{v} \right\|\cdot \left\| \mathbf{w} \right\|} \\ & =\frac{-32}{\sqrt{20}\cdot \sqrt{157}} \\ & =\frac{-32}{\sqrt{3140}} \end{align} Using a calculator we will get \begin{align} & \cos \theta =\frac{-32}{\sqrt{3140}} \\ & \theta ={{\cos }^{-1}}\left( \frac{-32}{\sqrt{3140}} \right) \\ & \theta \approx 124.8{}^\circ \end{align} Hence, $\mathbf{v}\cdot \mathbf{w}=-32$ and the angle between $\mathbf{v}$ and $\mathbf{w}$ is $\theta \approx 124.8{}^\circ$.

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