## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 792: 3

#### Answer

$\mathbf{v}\cdot \mathbf{w}=-6$ and $\mathbf{v}\cdot \mathbf{v}=41$

#### Work Step by Step

The dot product, $\mathbf{v}\cdot \mathbf{w}$ as, \begin{align} & \mathbf{v}\cdot \mathbf{w=}\left( 5\mathbf{i}-4\mathbf{j} \right)\cdot \left( -2\mathbf{i}-\mathbf{j} \right) \\ & =5\mathbf{i}\cdot \left( -2 \right)\mathbf{i+}5\mathbf{i}\cdot \left( -1 \right)\mathbf{j+}\left( -4 \right)\mathbf{j}\cdot \left( -2 \right)\mathbf{i+}\left( -4 \right)\mathbf{j}\cdot \left( -1 \right)\mathbf{j} \\ & =-10\left( \mathbf{i}\cdot \mathbf{i} \right)-5\left( \mathbf{i}\cdot \mathbf{j} \right)+8\left( \mathbf{j}\cdot \mathbf{i} \right)+4\left( \mathbf{j}\cdot \mathbf{j} \right) \\ & =-10\left( 1 \right)-5\left( 0 \right)+8\left( 0 \right)+4\left( 1 \right) \end{align} Solve ahead to get the result as, \begin{align} & \mathbf{v}\cdot \mathbf{w}=-10\left( 1 \right)-5\left( 0 \right)+8\left( 0 \right)+4\left( 1 \right) \\ & =-10+0+0+4 \\ & =-6 \end{align} The dot product, $\mathbf{v}\cdot \mathbf{v}$ as, \begin{align} & \mathbf{v}\cdot \mathbf{v=}\left( 5\mathbf{i}-4\mathbf{j} \right)\cdot \left( 5\mathbf{i}-4\mathbf{j} \right) \\ & =5\mathbf{i}\cdot 5\mathbf{i+}5\mathbf{i}\cdot \left( -4 \right)\mathbf{j+}\left( -4 \right)\mathbf{j}\cdot 5\mathbf{i+}\left( -4 \right)\mathbf{j}\cdot \left( -4 \right)\mathbf{j} \\ & =25\left( \mathbf{i}\cdot \mathbf{i} \right)-20\left( \mathbf{i}\cdot \mathbf{j} \right)-20\left( \mathbf{j}\cdot \mathbf{i} \right)+16\left( \mathbf{j}\cdot \mathbf{j} \right) \\ & =25\left( 1 \right)-20\left( 0 \right)-20\left( 0 \right)+16\left( 1 \right) \end{align} Solve ahead to get the result as, \begin{align} & \mathbf{v}\cdot \mathbf{v}=25\left( 1 \right)-20\left( 0 \right)-20\left( 0 \right)+16\left( 1 \right) \\ & =25+0+0+16 \\ & =41 \end{align}

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