Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 713: 29


Proof given below.

Work Step by Step

${\bf w_{1}}=\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}.$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}\cdot \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}= \frac{({\bf u}\cdot{\bf v})^{2}}{|{\bf v}|^{4}}\cdot|{\bf v}|^{2}= \frac{({\bf u}\cdot{\bf v})^{2}}{|{\bf v}|^{2}}.$ . ${\bf w_{2}}={\bf u}\displaystyle \cdot \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf u}\cdot{\bf v}= \frac{({\bf u}\cdot{\bf v})^{2}}{|{\bf v}|^{2}}.$ $( {\bf u} -\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u})\cdot \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}={\bf w_{1}-w_{2}}$ $\displaystyle \frac{({\bf u}\cdot{\bf v})^{2}}{|{\bf v}|^{2}}-\frac{({\bf u}\cdot{\bf v})^{2}}{|{\bf v}|^{2}}=0$
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