Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 14 / a

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.

For the triangular face, take the cross product of two edges and divide by two:The origin lies on the opposite side from this normal, since $(1,1,2)\cdot \overrightarrow{PO}=-3$. Thus this is already the outward vector area, withThe vector areas of a closed polyhedron sum to zero, so the sum for the remaining three faces is

Their individual outward vector areas are $(0,0,-3/2)$, $(1/2,-1,1)$, and $(-1,1/2,-1/2)$, which directly check that sum.

The normal from (ii) gives the plane equation $x+y+2z=3$. The distance from a point to a plane is the absolute value of the normalized plane equation at that point. At the origin this givesFor a geometric check, the perpendicular foot is $H=(1/2,1/2,1)$, parallel to the normal and satisfying the plane equation. Its Euclidean norm is the same distance.