Answer
(a) $i+j+2k$
(b) $\frac{\sqrt {6}}{6}i+ \frac{\sqrt {6}}{6}j+ \frac{\sqrt {6}}{3}k$
Work Step by Step
(a) Given $\vec u=i-j+0k$ and $\vec v=0i+2j-k$, by definition the cross product is given by $\vec u\times\vec v=\langle (-1\times(-1)-0\times2),(0\times0-1\times(-1)),(1\times2+1\times0) \rangle=\langle 1,1,2 \rangle= i+j+2k$
(b) Since the direction of the cross product is perpendicular to both $\vec u$ and $\vec v$, we can find a unit vector perpendicular to both $\vec u$ and $\vec v$ as $\vec n=\frac{1}{\sqrt {(1)^2+(1)^2+(2)^2}}\langle 1,1,2 \rangle=\langle \frac{\sqrt {6}}{6}, \frac{\sqrt {6}}{6}, \frac{\sqrt {6}}{3}\rangle=\frac{\sqrt {6}}{6}i+ \frac{\sqrt {6}}{6}j+ \frac{\sqrt {6}}{3}k$
