Chapter 12: Vectors and the Geometry of Space - Section 12.4 - The Cross Product - Exercises 12.4 - Page 719: 35

$2$

We have $\vec{AB}= \lt 0,1 \gt -\lt 1,0 \gt =\lt -1, 1 \gt$ and $\vec{AC}=\lt 0,-1 \gt -\lt 1,0 \gt =\lt -1, -1 \gt$ $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&0&0\\-1&-1&0\end{vmatrix}=0-0+(1+1)\hat{k}= 2\hat{k}$. Therefore, $|\vec{AB}\times\vec{AC}|=\sqrt {(2)^2}=2$