# Altitude of a right simplex (high-dimensional generalization of a 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

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 to altitude, which in 3-dimensions would be the length from the origin to the closest point on the face opposite the vertex. In $n$-dimensions it is difficult to explain the shape that is created.

Is the formula simply the following? Or how would one derive the formula?

$f = \frac{\prod_{i=0}^{n}{l_i}}{\sqrt{\sum_{i=0}^n{l_i^2}}}$

I'll write $a_i$ rather than $l_i$ for the edge of the simplex along the $i$th axis, since I find that less confusing to read.

We're looking for the distance from the origin to the hyperplane $H$ that passes through the $n$ vertices $(a_1,0,\ldots,0),$ $(0,a_2,0,\ldots,0),$ $(0,0,a_3,0,\ldots,0),$ and so forth up to $(0,\ldots,0,a_n).$ That hyperplane has equation $$\frac{x_1}{a_1} + \frac{x_2}{a_2} + \frac{x_3}{a_3} + \cdots + \frac{x_n}{a_n} = 1.$$

This can also be written as an inner product of vectors: $$\begin{bmatrix} \frac1{a_1}& \frac1{a_2}& \frac1{a_3}& \cdots& \frac1{a_n} \end{bmatrix} \begin{bmatrix} x_1& x_2& x_3& \cdots& x_n \end{bmatrix}^T = 1,$$ which tells us that the vector $\newcommand{v}{\mathbf v}\v = \begin{bmatrix} \frac1{a_1}& \frac1{a_2}& \frac1{a_3}& \cdots& \frac1{a_n} \end{bmatrix}$ is orthogonal to hyperplane $H$ and is therefore parallel to the altitude.

Let $k\v$ be the vector from the origin to the nearest point on the hyperplane $H.$ Then its inner product with $\v$ is $\v (k\v)^T = 1.$ This implies that $k \|\v\|^2 = k(\v\v^T) = 1$ as well, so $$k = \frac1{\| \v \|^2} = \frac{1}{\frac1{a_1^2} + \frac1{a_2^2} + \frac1{a_3^2} + \cdots + \frac1{a_n^2}}$$ But then $$f = \|k\v\| = k\|\v\| = \frac1{\|\v\|} = \frac{1}{\sqrt{\frac1{a_1^2} + \frac1{a_2^2} + \frac1{a_3^2} + \cdots + \frac1{a_n^2}}}$$ is the distance from the origin to the hyperplane $H.$

Let's check this for the case $n=2.$ If $n=2$ then $$f = \frac{1}{\sqrt{\frac1{a_1^2} + \frac1{a_2^2}}} = \frac{1}{\sqrt{\frac{a_1^2 + a_2^2}{a_1^2 a_2^2}}} = \frac{a_1 a_2}{\sqrt{a_1^2 + a_2^2}},$$ which agrees with the two-dimensional formula.

For $n=3,$ however, we have $$f = \frac{1}{\sqrt{\frac1{a_1^2} + \frac1{a_2^2} + \frac1{a_3^2}}} = \frac{1}{\sqrt{\frac{a_1^2 a_2^2 + a_1^2 a_3^2 + a_2^2 a_3^2} {a_1^2 a_2^2 a_3^2}}} = \frac{a_1 a_2 a_3}{\sqrt{a_1^2 a_2^2 + a_1^2 a_3^2 + a_2^2 a_3^2}},$$ whose numerator matches the proposed formula, $$\frac{\displaystyle\prod_{i=0}^n a_i} {\sqrt{\displaystyle\sum_{i=0}^n a_i^2}},$$ but whose denominator does not match that formula. The altitude could be written as follows:

$$f = \frac{\displaystyle\prod_{i=0}^n a_i} {\sqrt{\displaystyle\sum_{i=0}^n \displaystyle\prod_{j\neq i} a_j^2}}.$$

But I think it's simpler to write

$$f = \frac{1} {\sqrt{\displaystyle\sum_{i=0}^n \frac1{a_i^2}}}.$$

• This is cool, because the answer is $\frac{1}{f^2} = \sum_{i=0}^{n}{\frac{1}{a_i^2}}$ which is a simple formula – Paul Terwilliger May 25 '17 at 17:31