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: 1


$a.\qquad {\bf u}\cdot{\bf v}=-25, \ |{\bf u}|=5, \ {|\bf v|}=5$ $b.\qquad -1$ $c.\qquad -5$ $d.\qquad -2{\bf i}+4{\bf j}- \sqrt{5}{\bf k}$

Work Step by Step

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