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

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.