## Thomas' Calculus 13th Edition

${\bf u}\times{\bf v}$ 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 $\displaystyle \frac{\sqrt{3}}{3}{\bf i}+\frac{\sqrt{3}}{3}{\bf j}-\frac{\sqrt{3}}{3}{\bf k}$
${\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3} \end{array}\right|$ $=(u_{2}v_{3}-u_{3}v_{2}){\bf i}-(u_{1}v_{3}-u_{3}v_{1}){\bf j}+(u_{1}v_{2}-u_{2}v_{1}){\bf k}$ --- ${\bf w}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 3/2 & -1/2 & 1\\ 1 & 1 & 2 \end{array}\right|$ $=(-1-1){\bf i}-(3-1){\bf j}+(3/2+1/2){\bf k}$ $=-2{\bf i}-2{\bf j}+2{\bf k}$ $|{\bf w}|=\sqrt{4+4+4}=\sqrt{12}=2\sqrt{3}$ 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 $\displaystyle \frac{\sqrt{3}}{3}{\bf i}+\frac{\sqrt{3}}{3}{\bf j}-\frac{\sqrt{3}}{3}{\bf k}$