## Thomas' Calculus 13th Edition

${\bf F}=( 1.5{\bf i}- 0.5{\bf j})+( 0.5{\bf i}+ 1.5{\bf j}-3{\bf j})$
${\bf F}=\langle 2,1,-3\rangle,\quad {\bf v}=\langle 3,-1,0\rangle$ 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 ${\bf v}$ $\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf F}=\frac{{\bf F}\cdot{\bf v}}{|{\bf v}|^{2}}\cdot{\bf v}=\frac{6-1+0}{(\sqrt{9+1+0})} \langle 3,-1,0\rangle$ $=0.5\langle 3,-1,0\rangle$ $= 1.5{\bf i}- 0.5{\bf j}$ ${\bf F}-\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf F} = (2-1.5){\bf i}+ (1+0.5){\bf j}+ (-3-0){\bf j}$ $= 0.5{\bf i}+1.5{\bf j}-3{\bf j}$ ${\bf F}=( 1.5{\bf i}- 0.5{\bf j})+( 0.5{\bf i}+ 1.5{\bf j}-3{\bf j})$