Answer
(a) $\langle 0,2,8 \rangle$
(b) $\langle 0, \frac{\sqrt {17}}{17}, \frac{4\sqrt {17}}{17} \rangle$
Work Step by Step
(a) Given $\vec u=\langle 2,3,0 \rangle$ and $\vec v=\langle 0,4,-1 \rangle$, by definition the cross product is given by $\vec u\times\vec v=\langle (3\times0-0\times4),(0\times0-2\times(-1)),(2\times4-3\times0) \rangle=\langle 0,2,8 \rangle$
(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 $\vec n=\frac{1}{\sqrt {(0)^2+(2)^2+(8)^2}}\langle 0,2,8 \rangle=\langle 0, \frac{\sqrt {17}}{17}, \frac{4\sqrt {17}}{17} \rangle$
