Precalculus (6th Edition) Blitzer

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

Chapter 6 - Review Exercises - Page 800: 102

Answer

The projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}=\frac{50}{41}\mathbf{i}+\frac{40}{41}\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=\frac{50}{41}\mathbf{i}+\frac{40}{41}\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=-\frac{132}{41}\mathbf{i}+\frac{165}{41}\mathbf{j}$.

Work Step by Step

The projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}$ can be obtained as below: $\begin{align} & \text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v=}\frac{\mathbf{v}\cdot \mathbf{w}}{{{\left| \mathbf{w} \right|}^{2}}}\mathbf{w} \\ & =\frac{\left( -2\mathbf{i}+5\mathbf{j} \right)\cdot \left( 5\mathbf{i}+4\mathbf{j} \right)}{{{\left( \sqrt{{{5}^{2}}+{{4}^{2}}} \right)}^{2}}}\left( 5\mathbf{i}+4\mathbf{j} \right) \\ & =\frac{-2\left( 5 \right)+5\left( 4 \right)}{{{\left( \sqrt{25+16} \right)}^{2}}}\left( 5\mathbf{i}+4\mathbf{j} \right) \\ & =\frac{-10+20}{{{\left( \sqrt{41} \right)}^{2}}}\left( 5\mathbf{i}+4\mathbf{j} \right) \end{align}$ We will solve ahead to get the result as, $\begin{align} & \text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}=\frac{10}{41}\left( 5\mathbf{i}+4\mathbf{j} \right) \\ & =\frac{50}{41}\mathbf{i}+\frac{40}{41}\mathbf{j} \end{align}$ Now, we will obtain ${{\mathbf{v}}_{\mathbf{1}}}$ such that ${{\mathbf{v}}_{\mathbf{1}}}$ is parallel to $\mathbf{w}$ as, $\begin{align} & {{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v} \\ & =\frac{50}{41}\mathbf{i}+\frac{40}{41}\mathbf{j} \end{align}$ Now we will obtain ${{\mathbf{v}}_{\mathbf{2}}}$ such that ${{\mathbf{v}}_{\mathbf{2}}}$ is orthogonal to $\mathbf{w}$ as, $\begin{align} & {{\mathbf{v}}_{\mathbf{2}}}=\mathbf{v}-{{\mathbf{v}}_{\mathbf{1}}} \\ & =\left( -2\mathbf{i}+5\mathbf{j} \right)-\left( \frac{50}{41}\mathbf{i}+\frac{40}{41}\mathbf{j} \right) \\ & =-2\mathbf{i}+5\mathbf{j}-\frac{50}{41}\mathbf{i}-\frac{40}{41}\mathbf{j} \\ & =-\frac{132}{41}\mathbf{i}+\frac{165}{41}\mathbf{j} \end{align}$ Hence, the projection vector, $\text{pro}{{\text{j}}_{\mathbf{w}}}\mathbf{v}=\frac{50}{41}\mathbf{i}+\frac{40}{41}\mathbf{j}$ and ${{\mathbf{v}}_{\mathbf{1}}}=\frac{50}{41}\mathbf{i}+\frac{40}{41}\mathbf{j}$, ${{\mathbf{v}}_{\mathbf{2}}}=-\frac{132}{41}\mathbf{i}+\frac{165}{41}\mathbf{j}$.
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