If there are $(n+1)$ points in $m$ dimensional space, and we have known the Euclidean distances from one point "$B$" to the other $n$ points "$A_1,\ldots,A_n$", and known the positions of these $n$ points "$A_1,\ldots,A_n$", how can we get the position of the point "$B$" in $m$ dimensional space?
I understand it will form Two times simultaneous equations, as below example, but what is a good mathematical method for solving it? (Assume the points are in higher dimensional space, where m > 3)
*Example:
Assume the poisition of B is ($X_1,\ldots,X_m$), and the distance equations with other n point are below. I want to solve ($X_1,\ldots,X_m$)
($X_1-2)^2+\ldots+(X_m-4)^2$=5
($X_1-6)^2+\ldots+(X_m-7)^2$=9
($X_1-9)^2+\ldots+(X_m-5)^2$=11
... (totally n equations)
I'll show you how it can be done for $m = 2$.
$$\begin{align} (x_1 - a_1)^2 + (x_2 - a_2)^2 & = d_a^2\\ (x_1 - b_1)^2 + (x_2 - b_2)^2 & = d_b^2\\ (x_1 - c_1)^2 + (x_2 - c_2)^2 & = d_c^2\\ \end{align}$$
When you multiply each equation out you get this:
$$x_1^2 - 2a_1x_1 + a_1^2 + x_2^2 - 2 a_2x_2 + a_2^2= d_a^2$$
Which can be simplified to something that looks like this: $$x_1^2 + k_1 x_1 + x_2^2 + k_2 x_2 = k_3$$
Put all the equations in this form, then subtract the first equation from the rest. These transformations have not effected the overall solution space of the equations.
$$\begin{align} x_1^2 + a_{11} x_1 + x_2^2 + a_{12} x_2 & = e_1 \\ a_{21} x_1 + a_{22} x_2 & = e_2\\ a_{31} x_1 + a_{32} x_2 & = e_3\\ \end{align}$$
If $a_{21}, a_{22}, a_{31}, a_{32}$ form a nonsingular matrix then you can solve the linear system and have a single point to test against the first equation.
The intuition here is that the intersection of two n+1-dimensional spheres will always be a subset of an n-dimensional plane. Two circles will always intersect along a line, and two spheres will always intersect along a plane.