## Thomas' Calculus 13th Edition

$a.\quad 1$ $b.\quad 1$
$a.\quad$ The area of the paralellogram equals $|{\bf u}\times{\bf v}|$ ${\bf u}\times{\bf v}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 1 & 0\\ 0 & 1 & 0 \end{array}\right|$ $=(0-0){\bf i}-(0-0){\bf j}+(1-0){\bf k}$ $={\bf k}$ Area = $|{\bf u}\times{\bf v}| = \sqrt{0+0+1}=1$ $b.\quad$ Volume = $({\bf u}\times{\bf v} )\cdot{\bf w}\qquad$(triple scalar product) We can calculate this as a determinant, but we already have ${\bf u}\times{\bf v}$, so Volume $=0(1)+0(1)+1(1)$ $=1$