Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 1

Write $\mathbf{r}=\mu \mathbf{n}+\mathbf{v}$. Since $\mathbf{n}$ is a unit vector, the equation becomes $\mathbf{v}\cdot \mathbf{n}=0$. Thus $\mathbf{v}$ ranges over the two-dimensional orthogonal complement of $\mathbf{n}$: the locus is an affine hyperplane through $\mu \mathbf{n}$, with unit normal $\mathbf{n}$. Conversely, every such vector satisfies the equation. Its perpendicular distance from the origin is $\boxed{|\mu |}$; $\mu$ is the signed distance in the chosen normal direction.

The normal vector must be parallel to $\mathbf{b}$, and the plane must contain $\mathbf{a}$. Therefore one choice isThe nonzero scalar triple product ensures $\mathbf{b}\ne 0$. Replacing both $\mathbf{n}$ and $\mu$ by their negatives describes the same affine hyperplane.

Two directions in the plane are $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$. Their cross product $\mathbf{N}=(\mathbf{b}-\mathbf{a})\times (\mathbf{c}-\mathbf{a})$ is a normal vector. Moreover,so $\mathbf{N}$ cannot vanish. The required unit normal and offset areTaking dot products with $\mathbf{a},\mathbf{b},\mathbf{c}$ gives the same offset for all three points.

This plane contains the direction $\mathbf{b}-\mathbf{a}$ and is parallel to $\mathbf{c}$. Put $\mathbf{N}=(\mathbf{b}-\mathbf{a})\times \mathbf{c}$. Its scalar triple product with $\mathbf{a}$ is again $D\ne 0$, proving $\mathbf{N}\ne 0$. HenceThe dot product with $\mathbf{c}$ vanishes, while the dot products with $\mathbf{a}$ and $\mathbf{b}$ equal $\mu$ after normalization. These verify both incidence and parallelism.