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: 3


$a.\qquad {\bf u}\cdot{\bf v}=25,\ |{\bf u}|=5,\ |{\bf v}|=15$ $b.\qquad 1/3$ $c.\qquad 5/3$ $d.\displaystyle \qquad \frac{10}{9}{\bf i}+ \frac{11}{9}{\bf j} - \frac{2}{9}{\bf k}$

Work Step by Step

${\bf u}=\langle 0, 3, 4\rangle \quad {\bf v}=\langle 10, 11, -2\rangle$ $a.$ ${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}=$ $=(0)(10)+(3)(11)+(4)(-2)$ $=33-8$ $=25$ $|{\bf u}|=\sqrt{(0)^{2}+(3)^{2}+(4)^{2}}=\sqrt{9+16}=5$ $|{\bf v}|=\sqrt{(10)^{2}+(11)^{2}+(-2)^{2}}=\sqrt{100+121+4}=15$ $b.$ $\displaystyle \cos\theta=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}=\frac{25}{(5)(15)}=\frac{1}{3}$ $c.$ $|{\bf u}|\displaystyle \cos\theta=5(\frac{1}{3})=\frac{5}{3}$ $d.$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}$ $=\displaystyle \frac{25}{15^{2}}\langle 10, 11, -2\rangle$ $=\displaystyle \frac{1}{9}\langle 10, 11, -2\rangle$ $= \displaystyle \frac{10}{9}{\bf i}+ \frac{11}{9}{\bf j} - \frac{2}{9}{\bf k}$
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