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 793: 42


The vector $\text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}-\mathbf{w} \right)$ is $0\mathbf{i}+0\mathbf{j}$.

Work Step by Step

The vector $\text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}-\mathbf{w} \right)$ can be obtained as, $\begin{align} & \text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}-\mathbf{w} \right)\mathbf{=}\frac{\left( \mathbf{v}-\mathbf{w} \right)\cdot \mathbf{u}}{{{\left| \mathbf{u} \right|}^{2}}}\mathbf{u} \\ & =\frac{\left[ \left( 3\mathbf{i}-2\mathbf{j} \right)-\left( -5\mathbf{j} \right) \right]\cdot \left( -\mathbf{i}+\mathbf{j} \right)}{{{\left( \sqrt{{{\left( -1 \right)}^{2}}+{{1}^{2}}} \right)}^{2}}}\left( -\mathbf{i}+\mathbf{j} \right) \\ & =\frac{\left( 3\mathbf{i}+3\mathbf{j} \right)\cdot \left( -\mathbf{i}+\mathbf{j} \right)}{\sqrt{{{\left( -1 \right)}^{2}}+{{1}^{2}}}}\left( -\mathbf{i}+\mathbf{j} \right) \\ & =\frac{3\cdot \left( -1 \right)+3\cdot 1}{2}\left( -\mathbf{i}+\mathbf{j} \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}+\mathbf{w} \right)=\frac{-3+3}{2}\left( -\mathbf{i}+\mathbf{j} \right) \\ & =0\left( -\mathbf{i}+\mathbf{j} \right) \\ & =0\mathbf{i}+0\mathbf{j} \end{align}$ Hence, the vector $\text{pro}{{\text{j}}_{\mathbf{u}}}\left( \mathbf{v}-\mathbf{w} \right)$ is $0\mathbf{i}+0\mathbf{j}$.
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