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: 2

Answer

$\mathbf{v}\cdot \mathbf{w}=15$ and $\mathbf{v}\cdot \mathbf{v}=18$

Work Step by Step

The dot product, $\mathbf{v}\cdot \mathbf{w}$ as, $\begin{align} & \mathbf{v}\cdot \mathbf{w=}\left( 3\mathbf{i+}3\mathbf{j} \right)\cdot \left( \mathbf{i+}4\mathbf{j} \right) \\ & =3\mathbf{i}\cdot \mathbf{i+}3\mathbf{i}\cdot 4\mathbf{j+}3\mathbf{j}\cdot \mathbf{i+}3\mathbf{j}\cdot 4\mathbf{j} \\ & =3\left( \mathbf{i}\cdot \mathbf{i} \right)+12\left( \mathbf{i}\cdot \mathbf{j} \right)+3\left( \mathbf{j}\cdot \mathbf{i} \right)+12\left( \mathbf{j}\cdot \mathbf{j} \right) \\ & =3\left( 1 \right)+12\left( 0 \right)+3\left( 0 \right)+12\left( 1 \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \mathbf{v}\cdot \mathbf{w}=3\left( 1 \right)+12\left( 0 \right)+3\left( 0 \right)+12\left( 1 \right) \\ & =3+0+0+12 \\ & =15 \end{align}$ The dot product, $\mathbf{v}\cdot \mathbf{v}$ as, $\begin{align} & \mathbf{v}\cdot \mathbf{v=}\left( 3\mathbf{i+}3\mathbf{j} \right)\cdot \left( 3\mathbf{i+}3\mathbf{j} \right) \\ & =3\mathbf{i}\cdot 3\mathbf{i+}3\mathbf{i}\cdot 3\mathbf{j+}3\mathbf{j}\cdot 3\mathbf{i+}3\mathbf{j}\cdot 3\mathbf{j} \\ & =9\left( \mathbf{i}\cdot \mathbf{i} \right)+9\left( \mathbf{i}\cdot \mathbf{j} \right)+9\left( \mathbf{j}\cdot \mathbf{i} \right)+9\left( \mathbf{j}\cdot \mathbf{j} \right) \\ & =9\left( 1 \right)+9\left( 0 \right)+9\left( 0 \right)+9\left( 1 \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \mathbf{v}\cdot \mathbf{v}=9\left( 1 \right)+9\left( 0 \right)+9\left( 0 \right)+9\left( 1 \right) \\ & =9+0+0+9 \\ & =18 \end{align}$
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