## Thomas' Calculus 13th Edition

$({\bf i}\times{\bf j})\times{\bf j}={\bf -i}$ ${\bf i}\times({\bf j}\times{\bf j})={\bf 0}$ The cross product is not associative.
${\bf i}\times{\bf j}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{array}\right|=(0-0){\bf i}-(0-0){\bf j}+(1-0){\bf k}$ $={\bf k}$ $({\bf i}\times{\bf j})\times{\bf j}={\bf k}\times{\bf j}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{array}\right|=(0-1){\bf i}-(0-0){\bf j}+(0-0){\bf k}$ $({\bf i}\times{\bf j})\times{\bf j}=-{\bf i}$ On the other hand, ${\bf(j\times j)}={\bf 0}$, and ${\bf i}\times({\bf j}\times{\bf j})={\bf i}\times{\bf 0}={\bf 0}$ The cross product is not associative. If it were associative, then it would be that $({\bf u}\times{\bf v}) \times{\bf w}={\bf u}\times({\bf v}\times{\bf w})$ and, in this case, this is not so.