Optimizing computation of distance to triangle using barycentric coordinates - distance

Question

Building on the discussions here and here. I'm trying to compute the shortest distance between a 3D line and a 3D triangle.
I'm using barycentric coordinates to determine whether or not the point is inside the triangle. So given a triangle defined by vertices UVW and a line defined by point AB, I first compute the intersection of line AB with the plane defined by UVW. Let's call this intersection P and assume I've already done the checks to verify whether or not the point actually intersects the plane at all.
I then compute barycentric coordinates (S,T) such that S is defined along the edge UV and T is defined along the edge UW. Naturally, if 0≤S and 0≤T and S+T≤1 then P is on the triangle (or its edge) and my distance to the triangle is obviously zero.
If that's not true then P is outside the triangle and I need to compute the distance. The guidance of from the first link says to project point P onto all three edges to get three candidate points. Adding those points to the three triangle's vertices, you then have six points to test against.
Isn't it easier than that, though? If T<0, then don't you already know that UV is the closest edge and you only have to test against the projection of P onto that line? Similarly, if S<0 then UW would be the closest edge. If T>0 and S>0 then VW is the closest edge.
Thus based on the signs of S and T you already know the closest edge and only have to compute the distance from P to its projection onto that edge. If the projection isn't inside the triangle, then the closest point is either vertex. Thus your computations are about 1/3 of the proposed methods.
Am I missing something here, or is this a valid optimization? I'm fairly new to barycentric coordinates and their attributes.

Answer 1

It turns out that the problem of closest distance from a point and from a line are very similar and can both be reduced to a pure 2D problem.
Distance from a point
By Pythagoras, the squared distance from a point to a point of the triangle is the sum of the squared distance to the plane of support of the triangle and the squared distance of the projection of the point to that plane.
The latter distance is precisely the distance from the normal line to the triangle.
Distance from a line
Looking in the direction of the line, you see the projected triangle and the line is reduced to a single point. The requested 3D distance is equal to the 2D distance seen on the projection.
To obtain the desired coordinates, you use an auxiliary coordinate frame such that Z is in the direction of the line (and XY is a perpendicular plane); for convenience, choose the origin of the new frame to be on the line. Then by just ignoring Z, you get a planar problem in XY. The change of coordinates is an affine tranformation.
Point vs. triangle
Consider the three triangles formed by the origin (projection of the point/line) and a pair of triangle vertices (taken in cyclic order). The signed area of these triangles is a mere 2x2 determinant.
If the three areas have the same sign, the point is inside. Otherwise, the signs tell you where you are among the six surrounding regions, either past an edge or past a vertex.
On the upper figure, the point is inside (three positive areas). On the other figure, it is outside of the top-right edge (one negative area). Also note that dividing an area by the length of the corresponding side, you get the distance to the side. (Factor 2 omitted.)
The total work is
compute the affine frame;
convert the coordinates of the 3 or 4 points;
compute the three signed areas;
if inside, you are done;
otherwise, if in an edge region, compute a distance to a line and two distances to points;
otherwise you are in a vertex region, compute two distances to lines and one distance to vertex.