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: 36

Answer

The $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ is $-\frac{6}{5}\mathbf{i}+\frac{12}{5}\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=-\frac{6}{5}\mathbf{i}+\frac{12}{5}\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=\frac{16}{5}\mathbf{i}+\frac{8}{5}\mathbf{j}$.

Work Step by Step

The projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ can be obtained as, $\begin{align} & \text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v=}\frac{\mathbf{v}\cdot \mathbf{w}}{{{\left| \mathbf{w} \right|}^{2}}}\mathbf{w} \\ & =\frac{\left( 2\mathbf{i}+4\mathbf{j} \right)\cdot \left( -3\mathbf{i}+6\mathbf{j} \right)}{{{\left( \sqrt{{{\left( -3 \right)}^{2}}+{{6}^{2}}} \right)}^{2}}}\left( -3\mathbf{i}+6\mathbf{j} \right) \\ & =\frac{2\cdot \left( -3 \right)+4\cdot 6}{45}\left( -3\mathbf{i}+6\mathbf{j} \right) \\ & =\frac{-6+24}{45}\left( -3\mathbf{i}+6\mathbf{j} \right) \end{align}$ Solve ahead to get the result as, $\begin{align} & \text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}=\frac{-6+24}{45}\left( -3\mathbf{i}+6\mathbf{j} \right) \\ & =\frac{18}{45}\left( -3\mathbf{i}+6\mathbf{j} \right) \\ & =-\frac{6}{5}\mathbf{i}+\frac{12}{5}\mathbf{j} \end{align}$ Now, obtain ${{\mathbf{v}}_{\mathbf{1}}}$ such that ${{\mathbf{v}}_{\mathbf{1}}}$ is parallel to $\mathbf{w}$ as, $\begin{align} & {{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v} \\ & =-\frac{6}{5}\mathbf{i}+\frac{12}{5}\mathbf{j} \end{align}$ Obtain ${{\mathbf{v}}_{\mathbf{2}}}$ such that ${{\mathbf{v}}_{\mathbf{2}}}$ is orthogonal to $\mathbf{w}$ as, $\begin{align} & {{\mathbf{v}}_{\mathbf{2}}}=\mathbf{v}-{{\mathbf{v}}_{\mathbf{1}}} \\ & =\left( 2\mathbf{i}+4\mathbf{j} \right)-\left( -\frac{6}{5}\mathbf{i}+\frac{12}{5}\mathbf{j} \right) \\ & =2\mathbf{i}+4\mathbf{j}+\frac{6}{5}\mathbf{i}-\frac{12}{5}\mathbf{j} \\ & =\frac{16}{5}\mathbf{i}+\frac{8}{5}\mathbf{j} \end{align}$ Hence, the projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ is $-\frac{6}{5}\mathbf{i}+\frac{12}{5}\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=-\frac{6}{5}\mathbf{i}+\frac{12}{5}\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=\frac{16}{5}\mathbf{i}+\frac{8}{5}\mathbf{j}$.
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