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 794: 84


See the explanation below.

Work Step by Step

Let vector $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$. The projection of $\mathbf{v}$ onto $\mathbf{i}$ is expressed as follows: $\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=\frac{\mathbf{v}\centerdot \mathbf{i}}{{{\left\| \mathbf{i} \right\|}^{2}}}\mathbf{i}$ (1) Substituting the value of $\mathbf{v}$ in equation (1) for $\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}$, we get $\begin{align} & \text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=\frac{\left( a\mathbf{i}+b\mathbf{j} \right)\centerdot \mathbf{i}}{{{\left\| \mathbf{i} \right\|}^{2}}}\mathbf{i} \\ & =\frac{\left( a\mathbf{i}+b\mathbf{j} \right)\centerdot \left( \mathbf{i}+0\mathbf{j} \right)}{{{\left( \sqrt{{{1}^{2}}+{{0}^{2}}} \right)}^{2}}}\left( \mathbf{i}+0\mathbf{j} \right) \\ & =\frac{a\left( 1 \right)+b\left( 0 \right)}{1}\mathbf{i} \end{align}$ That is, $\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=a\mathbf{i}$ (2) Now, calculate $\left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}$ to get $\begin{align} & \left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}=\left( \left( a\mathbf{i}+b\mathbf{j} \right)\centerdot \mathbf{i} \right)\mathbf{i} \\ & =\left( a\left( 1 \right)+b\left( 0 \right) \right)\mathbf{i} \end{align}$ That is, $\left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}=a\mathbf{i}$ (3) Comparing equations (2) and (3), we get $\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=\left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}$ Hence, it is proved that $\text{pro}{{\text{j}}_{\mathbf{i}}}\mathbf{v}=\left( \mathbf{v}\centerdot \mathbf{i} \right)\mathbf{i}$.
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