Answer
${\bf u}\times{\bf v}$ has length $2\sqrt{3}$ and direction$ -\displaystyle \frac{\sqrt{3}}{3}{\bf i}-\frac{\sqrt{3}}{3}{\bf j}+\frac{\sqrt{3}}{3}{\bf k}$
${\bf v}\times{\bf u}$ has length $2\sqrt{3}$ and direction $ \displaystyle \frac{\sqrt{3}}{3}{\bf i}+\frac{\sqrt{3}}{3}{\bf j}-\frac{\sqrt{3}}{3}{\bf k}$
Work Step by Step
${\bf u}\times{\bf v}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
u_{1} & u_{2} & u_{3}\\
v_{1} & v_{2} & v_{3}
\end{array}\right|$
$=(u_{2}v_{3}-u_{3}v_{2}){\bf i}-(u_{1}v_{3}-u_{3}v_{1}){\bf j}+(u_{1}v_{2}-u_{2}v_{1}){\bf k}$
---
${\bf w}={\bf u}\times{\bf v}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
3/2 & -1/2 & 1\\
1 & 1 & 2
\end{array}\right|$
$=(-1-1){\bf i}-(3-1){\bf j}+(3/2+1/2){\bf k}$
$=-2{\bf i}-2{\bf j}+2{\bf k}$
$|{\bf w}|=\sqrt{4+4+4}=\sqrt{12}=2\sqrt{3}$
and the unit vector parallel to ${\bf w} $is
$\displaystyle \frac{{\bf w} }{|{\bf w} |}= \frac{-2}{2\sqrt{3}}{\bf i}-\frac{2}{2\sqrt{3}}{\bf j}+\frac{2}{2\sqrt{3}}{\bf k}$
${\bf w}=2\displaystyle \sqrt{3}( -\frac{\sqrt{3}}{3}{\bf i}-\frac{\sqrt{3}}{3}{\bf j}+\frac{\sqrt{3}}{3}{\bf k})$
${\bf v}\displaystyle \times{\bf u}=-{\bf w}=2\sqrt{3}( \frac{\sqrt{3}}{3}{\bf i}+\frac{\sqrt{3}}{3}{\bf j}-\frac{\sqrt{3}}{3}{\bf k})$
${\bf u}\times{\bf v}$ has length $2\sqrt{3}$ and direction$ -\displaystyle \frac{\sqrt{3}}{3}{\bf i}-\frac{\sqrt{3}}{3}{\bf j}+\frac{\sqrt{3}}{3}{\bf k}$
${\bf v}\times{\bf u}$ has length $2\sqrt{3}$ and direction $ \displaystyle \frac{\sqrt{3}}{3}{\bf i}+\frac{\sqrt{3}}{3}{\bf j}-\frac{\sqrt{3}}{3}{\bf k}$
