Thomas' Calculus 13th Edition

${\bf i}\times{\bf (i+j)}={\bf k}$
${\bf i}\times{\bf (i+j)}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 0 & 0\\ 1 & 1 & 0 \end{array}\right|$ $=(0-0){\bf i}-(0-0){\bf j}+(1-0){\bf k}={\bf k}$