Answer
(a) $-2j$
(b) $-j$
Work Step by Step
(a) Given $\vec u=i+j-k$ and $\vec v=i-j+k$, by definition the cross product is given by $\vec u\times\vec v=\langle (1\times1+1\times(-1)),(-1\times1-1\times1),(1\times(-1)-1\times1) \rangle=\langle 0,-2,0 \rangle= 0i-2j+0k$
(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 $\vec n=\frac{1}{\sqrt {(0)^2+(-2)^2+(0)^2}}\langle 0,-2,0 \rangle=\langle 0, -1, 0\rangle=0i-j+0k$
