## Thomas' Calculus 13th Edition

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