Answer
$a.\qquad {\bf u}\cdot{\bf v}=\sqrt{3}-\sqrt{2},\ |{\bf u}|=3,\ |{\bf v}|=\sqrt{2}$
$b.\displaystyle \qquad \frac{\sqrt{6}-2}{6}$
$c.\displaystyle \qquad \frac{\sqrt{6}-2}{2}$
$d.\displaystyle \qquad \frac{\sqrt{2}-\sqrt{3}}{2}{\bf i} + \frac{\sqrt{3}-\sqrt{2}}{2}{\bf j}$
Work Step by Step
${\bf u}=\langle\sqrt{2},\sqrt{3},2\rangle \quad {\bf v}=\langle-1,1,0\rangle$
${\bf (a)}$
${\bf u}\cdot{\bf v}=u_{1}v_{1}+u_{2}v_{2}+u_{3}v_{3}=$
$=(\sqrt{2})(-1)+(\sqrt{3})(1)+(2)(0)$
$=\sqrt{3}-\sqrt{2}$
$|{\bf u}|=\sqrt{(\sqrt{2})^{2}+(\sqrt{3})^{2}+(2)^{2}}=\sqrt{2+3+4}=3$
$|{\bf v}|=\sqrt{(-1)^{2}+(1)^{2}+(0)^{2}}=\sqrt{2}$
${\bf (b)}$
$\displaystyle \cos\theta=\frac{{\bf u}\cdot{\bf v}}{|{\bf u}||{\bf v}|}=\frac{\sqrt{3}-\sqrt{2}}{(3)(\sqrt{2})}=\frac{\sqrt{6}-2}{6}$
${\bf (c)}$
$|{\bf u}|\displaystyle \cos\theta=3(\frac{\sqrt{6}-2}{6})=\frac{\sqrt{6}-2}{2}$
${\bf (d)}$
$\displaystyle \mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}_{{\bf v}}{\bf u}=(\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|^{2}}){\bf v}$
$=\displaystyle \frac{\sqrt{3}-\sqrt{2}}{2}\langle-1,1,0\rangle$
$= \displaystyle \frac{\sqrt{2}-\sqrt{3}}{2}{\bf i} + \frac{\sqrt{3}-\sqrt{2}}{2}{\bf j}$
