Find the area of the triangle with the vertices \(A(1,-1,2)\text{,}\)\(B(2,0,-1)\text{,}\) and \(C(0,-2,3)\text{.}\) Note that these are the same points that we had in the previous example.
To find the area of the triangle, we first compute the area of the parallelogram built on \(\vec{AB}\) and \(\vec{AC}\text{,}\) shown in Figure 4.16, and then divide by \(2\text{:}\)
\begin{align*}
\text{Area of the parallelogram} \amp=
\left|\vec{AB} \times \vec{AC}\right| = \sqrt 8 = 2\sqrt 2\\
\text{Area of the triangle} \amp=
\frac12 \left|\vec{AB} \times \vec{AC}\right| = \sqrt 2.
\end{align*}
