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: 6


$a.\qquad {\bf u}\cdot{\bf v}=\sqrt{3}-\sqrt{2},\ |{\bf u}|=3,\ |{\bf v}|=\sqrt{2}$ $b.\displaystyle \qquad \frac{\sqrt{6}-2}{6} $ $c.\displaystyle \qquad \frac{\sqrt{6}-2}{2}$ $d.\displaystyle \qquad \frac{\sqrt{2}-\sqrt{3}}{2}{\bf i} + \frac{\sqrt{3}-\sqrt{2}}{2}{\bf j}$

Work Step by Step

${\bf u}=\langle\sqrt{2},\sqrt{3},2\rangle \quad {\bf v}=\langle-1,1,0\rangle$ ${\bf (a)}$ ${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}=$ $=(\sqrt{2})(-1)+(\sqrt{3})(1)+(2)(0)$ $=\sqrt{3}-\sqrt{2}$ $|{\bf u}|=\sqrt{(\sqrt{2})^{2}+(\sqrt{3})^{2}+(2)^{2}}=\sqrt{2+3+4}=3$ $|{\bf v}|=\sqrt{(-1)^{2}+(1)^{2}+(0)^{2}}=\sqrt{2}$ ${\bf (b)}$ $\displaystyle \cos\theta=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}=\frac{\sqrt{3}-\sqrt{2}}{(3)(\sqrt{2})}=\frac{\sqrt{6}-2}{6}$ ${\bf (c)}$ $|{\bf u}|\displaystyle \cos\theta=3(\frac{\sqrt{6}-2}{6})=\frac{\sqrt{6}-2}{2}$ ${\bf (d)}$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}$ $=\displaystyle \frac{\sqrt{3}-\sqrt{2}}{2}\langle-1,1,0\rangle$ $= \displaystyle \frac{\sqrt{2}-\sqrt{3}}{2}{\bf i} + \frac{\sqrt{3}-\sqrt{2}}{2}{\bf j}$
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