Chapter 11 - Section 11.4 - The Cross Product - Exercises - Page 623: 42

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We have $\vec{AB}= 4\hat{i}+4\hat{j}$ and $\vec{AC}= 3\hat{i}+2\hat{j}$. Area of the triangle = $\frac{1}{2}|\vec{AB}\times\vec{AC}|$ Here, $\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}$. Therefore, $|\vec{AB}\times\vec{AC}|=\sqrt {16}= 4$ Area= $\frac{1}{2}\times4=2$