Finding vertex of a 3D Triangle on a Plane

[Anand Sivaram]

TL;DR Summary

3D Triangle third vertex on a Plane

I came across the following problem and wondering how to solve it.

There is a plane n1x + n2y + n3z + n4 = 0 where n1, n2, n3, n4 are known. The triangle is in this plane.
We already know the two vertices P1(x1, y1, z1), P2(x2, y2, z2) of the triangle.
Now we have to find the third vertex P(x, y, z) of the triangle on the plane
such that P1-P distance is L1 and P2-P distance is L2 and are known.

I tried to find the solution in a number of places and came across the following one, but I was wondering whether we could get a unique solution based on that. Because this solutions reaches the long equation and the plane equation already, that means two equations and three unknowns.
https://math.stackexchange.com/ques...le-with-known-plane-two-points-and-lengths-of

Is it possible to get a unique or two point solution for P? Any help would be really appreciated.

 

[jedishrfu]

I think the answer is there are two solutions as you surmised from a simple geometric argument using a compass and straightedge on a piece of paper. With the paper acting as the plane, draw a line segment on the paper as one edge of the triangle.

Drawing arcs centered on the endpoints of the line segment with the lengths L1 and L2. You will find two points, both solutions to your problem.

 

[Mark44]

Is it possible to get a unique or two point solution for P?

Not a unique solution.

Drawing arcs centered on the endpoints of the line segment with the lengths L1 and L2. You will find two points, both solutions to your problem.

Right, and the two solutions for P are symmetric across the line segment $P_1P_2$.

 

[mathman]

There should be three equations for P. P is on the plane and the two distance equations.