## Thomas' Calculus 13th Edition

$$2$$
We have $\vec{AB}= 4\hat{i}+4\hat{j} \\ \vec{AC}= 3\hat{i}+2\hat{j}$ The area of the triangle is: $$A=\dfrac{1}{2}|\vec{AB}\times\vec{AC}|$$ Now, $$\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4&4&0\\3&2&0\end{vmatrix} \\=(8-12)\hat{k} \\= -4\hat{k}$$ and $|\vec{AB}\times\vec{AC}|=\sqrt {16}= 4$ Now, $Area= 2$