Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 1

Subtract the two position vectors to obtain the direction $\mathbf{b}-\mathbf{a}=(-3,-1,3)$. The affine line in a vector space through the points is thereforeThe values $t=0,1$ recover the two specified points.

A normal vector to the plane spanned by $\mathbf{a},\mathbf{b}$ is their cross product:Dividing by this Euclidean norm gives the two possible choices of unit normal,

The two edge vectors based at $A$ are $\mathbf{b}-\mathbf{a}=(-3,-1,3)$ and $\mathbf{c}-\mathbf{a}=(-3,2,2)$. Their cross product is $(-8,-3,-9)$, with squared Euclidean norm $64+9+81=154$. The triangle occupies half the corresponding parallelogram, so

Use the normal vector $\mathbf{N}=(-8,-3,-9)$ found from the edge vectors, rather than the normal vector to the origin-based linear span in part (b). The required plane satisfies $\mathbf{N}\cdot (\mathbf{r}-\mathbf{a})=0$. Since $\mathbf{N}\cdot \mathbf{a}=5$,After dividing by $|\mathbf{N}|=\sqrt{154}$, its equation has a unit normal. Its perpendicular distance from the origin is thus $\boxed{5/\sqrt{154}}$.