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: 5


$a.\qquad {\bf u}\cdot{\bf v}=2,\ |{\bf u}|=\sqrt{3},\ |{\bf v}|=\sqrt{34}$ $b.\displaystyle \qquad \frac{\sqrt{102}}{51} $ $c.\displaystyle \qquad \frac{\sqrt{34}}{17}$ $d.\displaystyle \qquad \frac{5}{17}{\bf j} -\frac{3}{17}{\bf k}$

Work Step by Step

${\bf u}=\langle 1,1,1\rangle \quad {\bf v}=\langle 0,5,-3\rangle$ ${\bf (a)}$ ${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}=$ $=(1)(0)+(1)(5)+(1)(-3)$ $=5-3$ $=2$ $|{\bf u}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}}=\sqrt{3}$ $|{\bf v}|=\sqrt{(0)^{2}+(5)^{2}+(-3)^{2}}=\sqrt{25+9}=\sqrt{34}$ ${\bf (b)}$ $\displaystyle \cos\theta=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}=\frac{2}{(\sqrt{3})(\sqrt{34})}$ $=\displaystyle \frac{2}{\sqrt{102}}=\frac{2\sqrt{102}}{102} = \frac{\sqrt{102}}{51} $ ${\bf (c)}$ $|{\bf u}|\displaystyle \cos\theta=\sqrt{3}(\frac{2}{(\sqrt{3})(\sqrt{34})})$ $=\displaystyle \frac{2}{\sqrt{34}}=\frac{2\sqrt{34}}{34}=\frac{\sqrt{34}}{17}$ ${\bf (d)}$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}$ $=\displaystyle \frac{2}{34}\langle 0,5,-3\rangle$ $=\displaystyle \langle 0,\frac{5}{17},-\frac{3}{17}\rangle$ $= \displaystyle \frac{5}{17}{\bf j} -\frac{3}{17}{\bf k}$
This answer is currently locked

Someone from the community is currently working feverishly to complete this textbook answer. Don’t worry, it shouldn’t be long.