= Solution
Use outward <normal vectors> to the two faces. The <cross products>
$$
\mathbf n_1=\overrightarrow{OP}\times\overrightarrow{OR}=(1,-2,2),\qquad
\mathbf n_2=\overrightarrow{OR}\times\overrightarrow{OQ}=(-2,1,-1)
$$
point away from the opposite vertices, since $\mathbf n_1\cdot\overrightarrow{OQ}=-3$ and $\mathbf n_2\cdot\overrightarrow{OP}=-3$. Their <Euclidean norms> are $3$ and $\sqrt6$, and their <dot product> is $-6$. The <interior dihedral angle of a convex polyhedron> is supplementary to the outward-normal angle, so
$$
\boxed{\cos\theta=-\frac{\mathbf n_1\cdot\mathbf n_2}{|\mathbf n_1||\mathbf n_2|}
=\frac2{\sqrt6}=\sqrt{\frac23}}.
$$
This also equals the cosine of the acute angle between the unoriented face planes. The outward-normal angle itself has cosine $-\sqrt{2/3}$; keeping that distinction avoids a sign ambiguity.
Back to article page
