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 712: 8

Answer

$a.\displaystyle \qquad {\bf u}\cdot{\bf v}=\frac{1}{6},\ |{\bf u}|=\frac{\sqrt{30}}{6},\ |{\bf v}|=\frac{\sqrt{30}}{6}$ $b.\qquad 1/5$ $c.\displaystyle \qquad \frac{\sqrt{30}}{30}$ $ d.\displaystyle \qquad \langle \frac{1}{5\sqrt{2}},\frac{1}{5\sqrt{3}}\rangle$

Work Step by Step

${\bf (a)}$ ${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}=$ $=(\displaystyle \frac{1}{\sqrt{2}} )( \frac{1}{\sqrt{2}} )+( \frac{1}{\sqrt{3}} )(- \frac{1}{\sqrt{3}} )=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$ $|{\bf u}|=\displaystyle \sqrt{( \frac{1}{\sqrt{2}} )^{2}+( -\frac{1}{\sqrt{3}} )^{2}}=\sqrt{\frac{5}{6}}=\frac{\sqrt{5}\cdot\sqrt{6}}{6}=\frac{\sqrt{30}}{6}$ $|{\bf v}|=\displaystyle \sqrt{( \frac{1}{\sqrt{2}} )^{2}+( \frac{1}{\sqrt{3}} )^{2}}=\frac{\sqrt{30}}{6}$ ${\bf (b)}$ $\displaystyle \cos\theta=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}=\frac{\frac{1}{6}}{(\frac{\sqrt{30}}{6})(\frac{\sqrt{30}}{6})}=\frac{6}{30}=\frac{1}{5}$ ${\bf (c)}$ $|{\bf u}|\displaystyle \cos\theta=\frac{\sqrt{30}}{6}(\frac{1}{5})=\frac{\sqrt{30}}{30}$ ${\bf (d)}$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}$ $=\displaystyle \frac{\frac{1}{6}}{\frac{30}{36}}\langle \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{3}}\rangle=\frac{1}{5}\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}}\rangle$ $= \displaystyle \langle \frac{1}{5\sqrt{2}},\frac{1}{5\sqrt{3}}\rangle$
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