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The acute angle between two unoriented planes is the acute angle between their nonzero normal vectors, giving the displayed formula. For an oriented convex polyhedron, the interior dihedral angle of a convex polyhedron is instead $\pi$ minus the angle between outward face normals. The distinction matters when a face normal is reversed: its unoriented plane is unchanged, but its vector angle is replaced by its supplement.

Use outward normal vectors to the two faces. The cross productspoint 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 $\sqrt{6}$, and their dot product is $-6$. The interior dihedral angle of a convex polyhedron is supplementary to the outward-normal angle, soThis 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.