## Thomas' Calculus 13th Edition

$({\bf u}\times{\bf v}) \cdot{\bf w}= ({\bf v}\times{\bf w}) \cdot {\bf u} = -7$ (the equality stands) $V=7$
${\bf u}\times{\bf v}=({\bf 2i+j})\times({\bf 2i-j+k})=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 2 & 1 & 0\\ 2 & -1 & 1 \end{array}\right|$ $=(1-0){\bf i}-(2-0){\bf j}+(-2-2){\bf k}= {\bf i} -2{\bf j} -4{\bf k}$ $=\langle 1, -2, -4 \rangle$ ${\bf w}={\bf i+2k} = \langle 1, 0, 2 \rangle$ $({\bf u}\times{\bf v}) \cdot {\bf w}=1(1)-2(0)-4(2)= -7$ ${\bf v}\times{\bf w}=({\bf 2i-j+k})\times({\bf i+2k})=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 2 & -1 & 1\\ 1 & 0 & 2 \end{array}\right|$ $=(-2-0){\bf i}-(4-1){\bf j}+(0+1){\bf k}= -2{\bf i} -3{\bf j}+{\bf k}$ $=\langle-2, -3, 1 \rangle$ ${\bf u}= \langle 2, 1, 0 \rangle$ $({\bf v}\times{\bf w}) \cdot {\bf u}=-2(2)-3(1)+1(0)= -7$ Thus, $({\bf u}\times{\bf v}) \cdot{\bf w}= ({\bf v}\times{\bf w}) \cdot {\bf u} = -7.$ and the volume of the parallelepiped determined by the three vectors is $V=|({\bf u}\times{\bf v}) \cdot {\bf w}|=7$