# Normal vector for altitude of a high-dimensional right triangle?

Assume that I have a right triangle, where the right angle is centered at the origin, and the two sides go in the $x$ and $y$ directions. I can calculate the altitude of the triangle using the following formula:

$f = \frac{ab}{c} = \frac{ab}{\sqrt{a^2 + b^2}}$

Where $a$ and $b$ are the lengths of the sides of the triangle. $c$ is the length of the hypotenuse. $f$ is the length of the altitude.

The formula is obtained from here

My question in 2-dimensions is this: How do I find the unit vector pointing in the direction of the altitude?

Assume that I generalize this right triangle to higher dimensions. In 3-dimensions, this would be a tetrahedron with one vertex at the origin. 3 edges of the tetrahedron would lie on the $x$-axis, $y$-axis, and $z$-axis respectively. In higher dimensions, the pattern would stay the same.

In $n$-dimensions, I would like to find the generalization of the unit vector pointing in the direction of the altitude, which in 3-dimensions would be the unit vector pointing from the origin to the closest point on the face opposite the origin. In $n$-dimensions it is difficult to explain the shape that is created.

How do I find the unit vector pointing in the direction of altitude for an $n$-dimensional right simplex?

EDIT: I have asked a similar question here, but there is a difference. There, I am asking for the length of the altitude. Here, I am asking for the unit vector of the altitude.

The vertices of the simplex are $(0,0,\ldots,0)$, $(a_1,0,\ldots,0)$, $(0,a_2,\ldots,0),\ldots$ $(0,0,\ldots,a_n)$. The face opposite the origin has equation $x_1/a_1+\cdots+x_n/a_n=1$. One normal is $\mathbf{b}=(1/a_1,\ldots,1/a_n)$ and the unit normal is $|\mathbf{b}|^{-1}\mathbf{b}$.