Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 5

The coordinate intercepts suggest the normal vectorIndeed $\mathbf{n}\cdot \mathbf{a}=\mathbf{n}\cdot \mathbf{b}=\mathbf{n}\cdot \mathbf{c}=1$, so the two independent edge vectors in the plane are both orthogonal to $\mathbf{n}$.

The plane equation in normal form isAn equivalent vector parametrization is $\mathbf{r}=\mathbf{a}+s(\mathbf{b}-\mathbf{a})+t(\mathbf{c}-\mathbf{a})$, with $s,t\in \mathbb{R}$.

The distance from a point to a plane, applied at the origin, is the plane constant divided by the Euclidean norm of its normal vector. ThereforeThe closest point on a plane lies on the normal through the origin; this is an orthogonal projection onto an intercept plane.

Taking the dot product with $\mathbf{n}$ gives $\mathbf{n}\cdot \mathbf{r}=\lambda +\mu +\nu$. The necessary and sufficient condition is thereforeThese coefficients describe an affine combination of the three vertices, or their barycentric coordinate system. Nonnegativity would additionally restrict the point to the triangle; it is not required for the whole plane.

Put $S=a^{-2}+b^{-2}+c^{-2}$. The orthogonal projection onto an intercept plane is $\mathbf{d}=\mathbf{n}/S$. Comparing its coordinates with $(\lambda a,\mu b,\nu c)$ givesTheir sum is one, as the plane equation requires.

The vector $\mathbf{a}$ is orthogonal to both $\mathbf{b}$ and $\mathbf{c}$, so $\mathbf{a}\cdot (\mathbf{b}-\mathbf{c})=0$. Also $\mathbf{d}\cdot \mathbf{b}=\mathbf{d}\cdot \mathbf{c}=1/S$, hence $\mathbf{d}\cdot (\mathbf{b}-\mathbf{c})=0$. Subtraction gives $(\mathbf{a}-\mathbf{d})\cdot (\mathbf{b}-\mathbf{c})=0$. Both vectors are nonzero, soThis is the altitude property of the orthogonal projection onto an intercept plane: the projection point is the triangle's orthocenter.