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: 7


$a.\qquad {\bf u}\cdot{\bf v}=10+\sqrt{17},\ |{\bf u}|=\sqrt{21},\ |{\bf v}|=\sqrt{26}$ $b.\displaystyle \qquad \frac{10+\sqrt{17}}{\sqrt{546}}$ $c.\displaystyle \qquad \frac{10+\sqrt{17}}{\sqrt{26}}$ $d.\displaystyle \qquad \frac{5(10+\sqrt{17})}{26}{\bf i} + \frac{10+\sqrt{17}}{26}{\bf j}$

Work Step by Step

${\bf u}=\langle 2,\sqrt{17},0\rangle \quad {\bf v}=\langle 5,1,0\rangle$ ${\bf (a)}$ ${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}=$ $=(2)(5)+(\sqrt{17})(1)+(0)(0)$ $=10+\sqrt{17}$ $|{\bf u}|=\sqrt{(2)^{2}+(\sqrt{17})^{2}+(0)^{2}}=\sqrt{21}$ $|{\bf v}|=\sqrt{(5)^{2}+(1)^{2}+(0)^{2}}=\sqrt{26}$ ${\bf (b)}$ $\displaystyle \cos\theta=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}=\frac{10+\sqrt{17}}{(\sqrt{21})(\sqrt{26})}=\frac{10+\sqrt{17}}{\sqrt{546}}$ ${\bf (c)}$ $|{\bf u}|\displaystyle \cos\theta=\sqrt{21}(\frac{10+\sqrt{17}}{(\sqrt{21})(\sqrt{26})})=\frac{10+\sqrt{17}}{\sqrt{26}}$ ${\bf (d)}$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}$ $=\displaystyle \frac{10+\sqrt{17}}{26}\langle 5,1,0\rangle$ $= \displaystyle \frac{5(10+\sqrt{17})}{26}{\bf i} + \frac{10+\sqrt{17}}{26}{\bf j}$
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