Answer
$a.\qquad {\bf u}\cdot{\bf v}=13,\ |{\bf u}|=3,\ |{\bf v}|=15$
$b.\qquad 13/45$
$c.\qquad 13/15$
$d.\displaystyle \qquad \frac{26}{225}{\bf i}+ \frac{26}{45}{\bf j} - \frac{143}{225}{\bf k}$
Work Step by Step
${\bf u}=\langle 2,2,1\rangle \quad {\bf v}=\langle 2,10,-11\rangle$
$a.$
${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}=$
$=(2)(2)+(2)(10)+(1)(-11)$
$=4+20-11$
$=13$
$|{\bf u}|=\sqrt{(2)^{2}+(2)^{2}+(1)^{2}}=\sqrt{4+4+1}=3$
$|{\bf v}|=\sqrt{(2)^{2}+(10)^{2}+(-11)^{2}}=\sqrt{4+100+121}=15$
$b.$
$\displaystyle \cos\theta=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}=\frac{13}{(3)(15)}=\frac{13}{45}$
$c.$
$|{\bf u}|\displaystyle \cos\theta=3(\frac{13}{45})=\frac{13}{15}$
$d.$
$\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}$
$=\displaystyle \frac{13}{15^{2}}\langle 2,10,-11\rangle$
$=\displaystyle \frac{13}{225}\langle 2,10,-11\rangle$
$= \displaystyle \frac{26}{225}{\bf i}+ \frac{26}{45}{\bf j} - \frac{143}{225}{\bf k}$
