## Thomas' Calculus 13th Edition

${\bf u}\times{\bf v}$ has length $3$ and direction $\displaystyle \frac{2}{3}{\bf i}+\frac{1}{3}{\bf j}+\frac{2}{3}{\bf k}$ ${\bf v}\times{\bf u}$ has length $3$ and direction $-\displaystyle \frac{2}{3}{\bf i}-\frac{1}{3}{\bf j}-\frac{2}{3}{\bf k}$
${\bf u}\times{\bf v}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ u_{1} & u_{2} & u_{3}\\ v_{1} & v_{2} & v_{3} \end{array}\right|$ $=(u_{2}v_{3}-u_{3}v_{2}){\bf i}-(u_{1}v_{3}-u_{3}v_{1}){\bf j}+(u_{1}v_{2}-u_{2}v_{1}){\bf k}$ --- ${\bf w}={\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 2 & -2 & -1\\ 1 & 0 & -1 \end{array}\right|$ $=(2-0){\bf i}-(-2+1){\bf j}+(0+2){\bf k}$ $=2{\bf i}+{\bf j}+2{\bf k}$ $|{\bf w}|=\sqrt{4+1+4}=3$ and the unit vector parallel to ${\bf w}$is $\displaystyle \frac{{\bf w} }{|{\bf w} |}=\frac{2}{3}{\bf i}+\frac{1}{3}{\bf j}+\frac{2}{3}{\bf k}$ ${\bf w}=3(\displaystyle \frac{2}{3}{\bf i}+\frac{1}{3}{\bf j}+\frac{2}{3}{\bf k})$ ${\bf v}\displaystyle \times{\bf u}=-{\bf w}=3(-\frac{2}{3}{\bf i}-\frac{1}{3}{\bf j}-\frac{2}{3}{\bf k})$ ${\bf u}\times{\bf v}$ has length $3$ and direction $\displaystyle \frac{2}{3}{\bf i}+\frac{1}{3}{\bf j}+\frac{2}{3}{\bf k}$ ${\bf v}\times{\bf u}$ has length $3$ and direction $-\displaystyle \frac{2}{3}{\bf i}-\frac{1}{3}{\bf j}-\frac{2}{3}{\bf k}$