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 713: 30

Answer

${\bf F}=( 1.5{\bf i}- 0.5{\bf j})+( 0.5{\bf i}+ 1.5{\bf j}-3{\bf j})$

Work Step by Step

${\bf F}=\langle 2,1,-3\rangle,\quad {\bf v}=\langle 3,-1,0\rangle$ The projection of ${\bf F}$ onto ${\bf v}$ is a vector parallel to ${\bf v}$. Using the result of the previous exercise, ${\bf F}-\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf F}$ is orthogonal to ${\bf v}$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf F}=\frac{{\bf F}\cdot{\bf v}}{|{\bf v}|^{2}}\cdot{\bf v}=\frac{6-1+0}{(\sqrt{9+1+0})} \langle 3,-1,0\rangle$ $=0.5\langle 3,-1,0\rangle$ $= 1.5{\bf i}- 0.5{\bf j}$ ${\bf F}-\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf F} = (2-1.5){\bf i}+ (1+0.5){\bf j}+ (-3-0){\bf j}$ $= 0.5{\bf i}+1.5{\bf j}-3{\bf j}$ ${\bf F}=( 1.5{\bf i}- 0.5{\bf j})+( 0.5{\bf i}+ 1.5{\bf j}-3{\bf j})$
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