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: 4


$a.\qquad {\bf u}\cdot{\bf v}=13,\ |{\bf u}|=3,\ |{\bf v}|=15$ $b.\qquad 13/45$ $c.\qquad 13/15$ $d.\displaystyle \qquad \frac{26}{225}{\bf i}+ \frac{26}{45}{\bf j} - \frac{143}{225}{\bf k}$

Work Step by Step

${\bf u}=\langle 2,2,1\rangle \quad {\bf v}=\langle 2,10,-11\rangle$ $a.$ ${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}=$ $=(2)(2)+(2)(10)+(1)(-11)$ $=4+20-11$ $=13$ $|{\bf u}|=\sqrt{(2)^{2}+(2)^{2}+(1)^{2}}=\sqrt{4+4+1}=3$ $|{\bf v}|=\sqrt{(2)^{2}+(10)^{2}+(-11)^{2}}=\sqrt{4+100+121}=15$ $b.$ $\displaystyle \cos\theta=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}=\frac{13}{(3)(15)}=\frac{13}{45}$ $c.$ $|{\bf u}|\displaystyle \cos\theta=3(\frac{13}{45})=\frac{13}{15}$ $d.$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}$ $=\displaystyle \frac{13}{15^{2}}\langle 2,10,-11\rangle$ $=\displaystyle \frac{13}{225}\langle 2,10,-11\rangle$ $= \displaystyle \frac{26}{225}{\bf i}+ \frac{26}{45}{\bf j} - \frac{143}{225}{\bf k}$
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