Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 11

Choose Cartesian coordinates with unit vectors satisfying the right-hand rule, $\hat{\mathbf{x}}\times \hat{\mathbf{y}}=\hat{\mathbf{z}}$. Put $A$, $B$ and $C$ one unit along the positive $x$, $y$ and $z$ axes, respectively. The following original drawing also includes the vectors needed in the next part.

The position vectors end at the midpoints of $BC$, $CA$ and $AB$, respectively. The drawing places $E$, $F$ and $G$ on those edges and shows the arrows from the origin. The volume of their parallelepiped is the absolute scalar triple product:Equivalently, this is the absolute determinant of the matrix whose columns are the three vectors. Since it is nonzero, the vectors are linearly independent. Three linearly independent vectors in the three-dimensional vector space ${\mathbb{R}}^3$ form a basis.

Writing $\overrightarrow{OA}=a\mathbf{e}+b\mathbf{f}+c\mathbf{g}$ and comparing Cartesian coordinates givesThe unique solution, guaranteed by the basis property, is $(a,b,c)=(-1,1,1)$. Thus

Two directions within the plane areTheir cross product is $(1,1,1)/4$, so a unit vector normal to the plane isIts negative is equally valid. An equation of a plane through three points in vector form isor, parametrically, $\mathbf{r}=\mathbf{e}+s(\mathbf{f}-\mathbf{e})+t(\mathbf{g}-\mathbf{e})$. Since the coordinate sum of $E$ is one, the Cartesian equation of a plane isThe distance from a point to a plane is the absolute normal displacement, giving

Let $\mathbf{d}=\mathbf{f}-\mathbf{e}$. A point on the line has position vector $\mathbf{r}=\mathbf{e}+t\mathbf{d}$. By bilinearity and antisymmetry of the cross product,For a point with position vector $\mathbf{c}$, the area of the parallelogram spanned by $\mathbf{c}-\mathbf{e}$ and $\mathbf{d}$ equals base length times perpendicular Euclidean distance. Hence the shortest Euclidean distance to the line isFor $\mathbf{c}=(0,0,1)$, these quantities areThus $\boxed{d=\sqrt{6}/4}$. The perpendicular meets $EF$ at $(1/4,1/4,1/2)$, whose displacement from $C$ has precisely this length.