Let $H$ be a hyperplane defined by the points $p_1, p_2, ..., p_n$ and single point $x$ generally out of the hyperplane. Is there any formula to calculate the signed distance between $x$ and $H$?
I found, that following value is the orientation (i.e. its sign matches the sign of the required signed distance):
$\displaystyle \Delta_{H}(x)= \left\vert \begin{array}{ccccc} p_{1,1} & p_{1,2} & \cdots & p_{1,n} & 1 \\ p_{2,1} & p_{2,2} & \cdots & p_{2,n} & 1 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ p_{n,1} & p_{n,2} & \cdots & p_{n,n} & 1 \\ x_{1} & x_{2} & \cdots & x_{n} & 1 \end{array} \right\vert$
But I suspect, that $\Delta_H(x)$ have a bound with value of signed distance.
Am I right? What is the real formula of the signed distance from hyperplane to point?
Finally, I found the solution (in book Andrew J. Hanson: Geometry for N-Dimensional Graphics):
$ \displaystyle V=\frac{1}{n!} \left\vert \begin{array}{ccccc} p_{1,1} & p_{1,2} & \cdots & p_{1,n} & 1 \\ p_{2,1} & p_{2,2} & \cdots & p_{2,n} & 1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ p_{n,1} & p_{n,2} & \cdots & p_{n,n} & 1 \\ x_{1} & x_{2} & \cdots & x_{n} & 1 \notag \end{array} \right\vert \\ S=\frac{1}{(n - 1)!} \cdot \sqrt{ \left\vert \left ( \begin{array}{ccccc} p_{1,1} & p_{1,2} & \cdots & p_{1,n} & 1 \\ p_{2,1} & p_{2,2} & \cdots & p_{2,n} & 1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ p_{n,1} & p_{n,2} & \cdots & p_{n,n} & 1 \notag \end{array} \right ) \cdot \left ( \begin{array}{ccccc} p_{n,1} & p_{2,1} & \cdots & p_{1,1} \\ p_{n,2} & p_{2,2} & \cdots & p_{1,2} \\ \vdots & \vdots & \ddots & \vdots \\ p_{n,n} & p_{2,n} & \cdots & p_{1,n} \\ 1 & 1 & \cdots &1 \notag \end{array} \right ) \right\vert } \\ D=\frac{n\cdot V}{S} $