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

$\frac{11}{2}$
We have $\vec{AB}= -2\hat{i}+3\hat{j}$ and $\vec{AC}= 3\hat{i}+ \hat{j}$. The area of the given triangle = $\frac{1}{2}|\vec{AB}\times\vec{AC}|$ Now, $\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-2&3&0\\3&1&0\end{vmatrix}$$=(-2-9)\hat{k}= -11\hat{k}$. Therefore, $|\vec{AB}\times\vec{AC}|=\sqrt {121}= 11$. Thus the required area is $\frac{11}{2}$