Answer
\[
\frac{1}{6}(\vec{u} \cdot(\vec{w} \times \vec{v}) 1=v
\]
Work Step by Step
$V=\frac{1}{3}(\text { area of base })$ (height)
Area of base $=\|\vec{v} \times \vec{w}\|\frac{1}{2} \quad$ (area of triangle)
\[
\|\vec{u}\| \cos (\theta)=\text { height }
\]
The perpendicular from the vertex to the base is along the $\vec{w} \times \vec{v}$ direction. $\theta$ is the angle between $\vec{u} \times \vec{v}$ and $\vec{w}$
\[
\begin{array}{c}
V=\frac{1}{2} \cdot \frac{1}{3} \cdot\|\vec{w} \times \vec{v}\|\|\vec{u}\| \cos (\theta)=|\vec{u} \cdot(\vec{v} \times \vec{w})|\frac{1}{6} \\
\therefore \frac{1}{6}(\vec{u} \cdot(\vec{v} \times \vec{w}) 1=v
\end{array}
\]
