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

$\displaystyle \frac{3\sqrt{2}}{2}$
$\overrightarrow{AB} = \langle-1-0, 1-0,-1-0 \rangle= \langle-1, 1, -1 \rangle$ $\overrightarrow{AC} = \langle 3, 0, 3 \rangle$ We find the area of the parallelogram defined by these two vectors. The triangle has half the area of the parallelogram. $A=\displaystyle \frac{1}{2}| \overrightarrow{AB} \times \overrightarrow{AC} |$ $\overrightarrow{AB} \times \overrightarrow{AC}=\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ -1 & 1 & -1\\ 3 & 0 & 3 \end{array}\right|=\langle (3-0) ,\ -(-3+3) ,\ (0-3) \rangle$ $=\langle 3 ,\ 0 ,\ -3 \rangle$ $A=\displaystyle \frac{1}{2}\cdot\sqrt{9+0+9}=\frac{3\sqrt{2}}{2}$