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The common intersection of a triangle's three altitude lines. Each line passes through a vertex and is orthogonal to the opposite side. The intersection can lie inside or outside the triangle, or at its right-angle vertex. An orthogonal projection onto an intercept plane supplies a concrete three-dimensional construction of one.

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.

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.