## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 794: 65

#### Answer

The dot product of two vectors is $\mathbf{a}\cdot \mathbf{b}=\left\| a \right\|\text{ }\left\| b \right\|\cos \theta$.

#### Work Step by Step

Example: Information: Let vector $\mathbf{a}=2\mathbf{i}+\mathbf{j}+3\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+2\mathbf{j}+\mathbf{k}$, such that the angle between their directions $\theta ={{0}^{\circ }}$. Now, magnitude of vector a: \begin{align} & \left\| \mathbf{a} \right\|=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\ & =\sqrt{4+1+9} \\ & =\sqrt{14} \end{align} Therefore, $\left\| \mathbf{a} \right\|=\sqrt{14}$ Similarly, magnitude of vector b: \begin{align} & \left\| \mathbf{b} \right\|=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\ & =\sqrt{1+4+1} \\ & =\sqrt{6} \end{align} Therefore, $\left\| \mathbf{b} \right\|=\sqrt{6}$ Now, the cosine of the angle between their directions is $\cos \left( {{0}^{\circ }} \right)=1$ Hence, the dot product of the two vectors is \begin{align} & \mathbf{a}\cdot \mathbf{b}=\left( \sqrt{14} \right)\left( \sqrt{6} \right)\left( 1 \right) \\ & =9.165 \end{align}

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