Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 5

Let $\mathbf{p}=(2,1,1)$, $\mathbf{q}=(1,2,2)$ and $\mathbf{r}=(0,0,3)$. Cross products give plane normals $\mathbf{p}\times \mathbf{r}=(3,-6,0)$ and $\mathbf{q}\times \mathbf{r}=(6,-3,0)$. Their dot product is $36$ and each has norm $3\sqrt{5}$. Thus the acute angle between two planes isThis is also the interior dihedral angle here: perpendicular to the common edge $OR$, the interior wedge lies between the projected directions $(2,1)$ and $(1,2)$. The two outward face normals instead make the supplementary angle.

The face $PQR$ has edge vectors $\mathbf{q}-\mathbf{p}=(-1,1,1)$ and $\mathbf{r}-\mathbf{p}=(-2,-1,2)$. Their cross product is $(3,0,3)$, so choose the outward normal vector $\mathbf{n}=(1,0,1)$. The angle between vectors obeysReversing the chosen normal gives the supplementary angle $150^{\circ }$; the acute angle with the normal line is $30^{\circ }$.

The magnitude of the cross product of two edge vectors is the area of their parallelogram. The triangular face occupies half of it:

With normal vector $(1,0,1)$, the plane through $P$ has equation $x+z=3$, also satisfied by $Q,R$. Projecting the displacement from the origin onto the unit normal gives the distance from a point to a plane:The perpendicular foot is $(3/2,0,3/2)$, which lies in that plane and confirms the distance.