Recall "gravitational"-type force functions, by which I mean anything of the form:
$f(x,y,z) = \frac{k}{((x-x_0)^2+(y-y_0)^2+(z-z_0)^2)^p}, p\in\Re_{>0}, k\in\Re, (x,y,z) \neq(x_0,y_0,z_0)$
(e.g., for gravity, $p=1,k=G m_1 m_2$)
Define a function $g(x,y,z) = f_1(x,y,z) + f_2(x,y,z) + f_3(x,y,z) +\cdots+f_n(x,y,z)$ (a sum of several force functions, presumably each with different $k$ (though probably not $p$).
I have two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ such that $g(x_1,y_1,z_1) and $g(x_2,y_2,z_2)>a$. My problem is to find the location $(x_3,y_3,z_3)$ such that $g(x_3,y_3,z_3)=a$ where $(x_3,y_3,z_3)$ lies on the straight line between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$. If it helps, I know that exactly two of the xs, ys, and zs are equal.
I'm currently solving this numerically, but it will eventually be solved on an OpenCL kernel where I would very much like a closed form solution.
However, I am having trouble getting such a solution. I'm not even sure it's possible. I tried inverting $g$, but got basically nowhere. Ideas? Thanks.