## University Calculus: Early Transcendentals (3rd Edition)

$({\bf u}\times{\bf v}) \cdot{\bf w}= ({\bf v}\times{\bf w}) \cdot {\bf u} = 4$ (the equality is true) $V=4$
${\bf u}\times{\bf v}={\bf i-j+k}\times{\bf 2i+j-2k}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & -1 & 1\\ 2 & 1 & -2 \end{array}\right|$ $=(2-1){\bf i}-(-2-2){\bf j}+(1+2){\bf k}= {\bf i}+4{\bf j}+3{\bf k}$ $=\langle 1, 4, 3 \rangle$ ${\bf w}={\bf -i+2j-k} = \langle-1, 2, -1 \rangle$ $({\bf u}\times{\bf v}) \cdot {\bf w}=1(-1)+4(2)+3(-1)= 4$ ${\bf v}\times{\bf w}={\bf 2i+j-2k}\times{\bf -i+2j-k}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 2 & 1 & -2\\ -1 & 2 & -1 \end{array}\right|$ $=(-1+4){\bf i}-(-2-2){\bf j}+(4+1){\bf k}= 3{\bf i} +4{\bf j}+5{\bf k}$ $=\langle 3, 4, 5 \rangle$ ${\bf u}= \langle 1, -1, 1 \rangle$ $({\bf v}\times{\bf w}) \cdot {\bf u}=3(1)+4(-1)+5(1)=4.$ Thus, $({\bf u}\times{\bf v}) \cdot{\bf w}= ({\bf v}\times{\bf w}) \cdot {\bf u} = 4.$ and the volume of the parallelepiped determined by the three vectors is $V=|({\bf u}\times{\bf v}) \cdot {\bf w}|=4$