The area of a triangle with vertices at $\mathbf{u},$ $\mathbf{v},$ and $\mathbf{w}$ is given by
\[\frac{1}{2} \|\mathbf{v} - \mathbf{u} \| \|\mathbf{w} - \mathbf{u} \| \sin \theta,\]where $\theta$ is the angle between the vectors $\mathbf{v} - \mathbf{u}$ and $\mathbf{w} - \mathbf{u}.$
We can compute that
\begin{align*}
\mathbf{v} - \mathbf{u} &= (1, 3, 2) - (3,-2,-1) = (-2, 5, 3), \\
\mathbf{w} - \mathbf{u} &= (-2, 1, 3) - (3,-2,-1) = (-5, 3, 4).
\end{align*}Then
\[\|\mathbf{v} - \mathbf{u}\| = \sqrt{2^2 + 5^2 + 3^2} = \sqrt{38},\]and
\[\|\mathbf{w} - \mathbf{u}\| = \sqrt{(-5)^2 + 3^2 + 4^2} = \sqrt{50}.\]Furthermore,
\[\mathbf{v} - \mathbf{u} \cdot (\mathbf{w} - \mathbf{u}) = (-2)(-5) + 5 \cdot 3 + 3 \cdot 4 = 27,\]so
\[\sin \theta = \frac{27}{\sqrt{38} \sqrt{50}} = \frac{27}{10 \sqrt{19}}.\]Therefore, the area is
\[\frac{1}{2} \sqrt{38} \sqrt{50} \cdot \frac{27}{10 \sqrt{19}} = \boxed{\frac{27 \sqrt{190}}{10}}.\]
Find the area of the triangle having vertices at u = (3, -2, -1), v = (1, 3, 2), and w = (-2, 1, 3).
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