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: 2

Answer

$a.\qquad {\bf u}\cdot{\bf v}=3,\ |{\bf u}|=13,\ |{\bf v}|=1$ $b.\qquad 3/13$ $c.\qquad 3$ $d.\displaystyle \qquad \frac{9}{5}{\bf i}+ \frac{12}{5}{\bf k}$

Work Step by Step

${\bf u}=\langle 5, 12, 0\rangle \quad {\bf v}=\langle 3/5, 0, 4/5\rangle$ $a.$ ${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}=$ $=(5)(\displaystyle \frac{3}{5})+(12)(0)+(0)(4/5)$ $=3$ $|{\bf u}|=\sqrt{(5)^{2}+(12)^{2}+(0)^{2}}=\sqrt{25+144}=13$ $|{\bf v}|=\sqrt{(3/5)^{2}+(0)^{2}+(4/5)^{2}}=\sqrt{\frac{9+16}{25}}=1$ $b.$ $\displaystyle \cos\theta=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}=\frac{3}{(1)(13)}=\frac{3}{13}$ $c.$ $|{\bf u}|\displaystyle \cos\theta=13(\frac{3}{13})=1$ $d.$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}$ $=\displaystyle \frac{3}{1^{2}}\langle 3/5, 0, 4/5\rangle$ $=\langle 9/5, 0, 12/5\rangle$ $= \displaystyle \frac{9}{5}{\bf i}+ \frac{12}{5}{\bf k}$
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