Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 792: 4

Answer

$\mathbf{v}\cdot \mathbf{w}=-19$ and $\mathbf{v}\cdot \mathbf{v}=53$

Work Step by Step

The dot product, $\mathbf{v}\cdot \mathbf{w}$ as, $\begin{align} & \mathbf{v}\cdot \mathbf{w=}\left( 7\mathbf{i}-2\mathbf{j} \right)\cdot \left( -3\mathbf{i}-\mathbf{j} \right) \\ & =7\mathbf{i}\cdot \left( -3 \right)\mathbf{i+}7\mathbf{i}\cdot \left( -1 \right)\mathbf{j+}\left( -2 \right)\mathbf{j}\cdot \left( -3 \right)\mathbf{i+}\left( -2 \right)\mathbf{j}\cdot \left( -1 \right)\mathbf{j} \\ & =-21\left( \mathbf{i}\cdot \mathbf{i} \right)-7\left( \mathbf{i}\cdot \mathbf{j} \right)+6\left( \mathbf{j}\cdot \mathbf{i} \right)+2\left( \mathbf{j}\cdot \mathbf{j} \right) \\ & =-21\left( 1 \right)-7\left( 0 \right)+6\left( 0 \right)+2\left( 1 \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \mathbf{v}\cdot \mathbf{w}=-21\left( 1 \right)-7\left( 0 \right)+6\left( 0 \right)+2\left( 1 \right) \\ & =-21+0+0+2 \\ & =-19 \end{align}$ The dot product, $\mathbf{v}\cdot \mathbf{v}$ as, $\begin{align} & \mathbf{v}\cdot \mathbf{v=}\left( 7\mathbf{i}-2\mathbf{j} \right)\cdot \left( 7\mathbf{i}-2\mathbf{j} \right) \\ & =7\mathbf{i}\cdot 7\mathbf{i+}7\mathbf{i}\cdot \left( -2 \right)\mathbf{j+}\left( -2 \right)\mathbf{j}\cdot 7\mathbf{i+}\left( -2 \right)\mathbf{j}\cdot \left( -2 \right)\mathbf{j} \\ & =49\left( \mathbf{i}\cdot \mathbf{i} \right)-14\left( \mathbf{i}\cdot \mathbf{j} \right)-14\left( \mathbf{j}\cdot \mathbf{i} \right)+4\left( \mathbf{j}\cdot \mathbf{j} \right) \\ & =49\left( 1 \right)-14\left( 0 \right)-14\left( 0 \right)+4\left( 1 \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \mathbf{v}\cdot \mathbf{v}=49\left( 1 \right)-14\left( 0 \right)-14\left( 0 \right)+4\left( 1 \right) \\ & =49+0+0+4 \\ & =53 \end{align}$
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