## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 793: 37

#### Answer

The $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ is $\mathbf{i}+2\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=\mathbf{i}+2\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=0\mathbf{i}+0\mathbf{j}$.

#### 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( \mathbf{i}+2\mathbf{j} \right)\cdot \left( 3\mathbf{i}+6\mathbf{j} \right)}{{{\left( \sqrt{{{3}^{2}}+{{6}^{2}}} \right)}^{2}}}\left( 3\mathbf{i}+6\mathbf{j} \right) \\ & =\frac{1\cdot 3+2\cdot 6}{45}\left( 3\mathbf{i}+6\mathbf{j} \right) \\ & =\frac{3+12}{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{3+12}{45}\left( 3\mathbf{i}+6\mathbf{j} \right) \\ & =\frac{15}{45}\left( 3\mathbf{i}+6\mathbf{j} \right) \\ & =\frac{1}{3}\left( 3\mathbf{i}+6\mathbf{j} \right) \\ & =\mathbf{i}+2\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} \\ & =\mathbf{i}+2\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( \mathbf{i}+2\mathbf{j} \right)-\left( \mathbf{i}+2\mathbf{j} \right) \\ & =\mathbf{i}+2\mathbf{j}-\mathbf{i}-2\mathbf{j} \\ & =0\mathbf{i}+0\mathbf{j} \end{align} Hence, the projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ is $\mathbf{i}+2\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=\mathbf{i}+2\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=0\mathbf{i}+0\mathbf{j}$.

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