What is, and is there, a fast way to check where in the plane my line will intersect, if i know the plane is always in the same z-axis (so it cannot be rotated), and its width/height is infinite? Also, my "line" isn't actually a line, but a 3d vector, so the "line" can go to infinite distance.
Here is the code that relies on two points: (p1 and p2 are start and end points of the line. plane_z = where the plane is)
k1 = -p2.z/(p1.z-p2.z-plane_z);
k2 = 1.0f-k1开发者_运维问答;
ix = k1*p1.x + k2*p2.x;
iy = k1*p1.y + k2*p2.y;
iz = plane_z; // where my plane lays
Another solution which works with a vector (i made it use two points as the first example did too, "p2.x-p1.x" etc. is the vector calculation):
a = (plane_z-p1.z)/(p2.z-p1.z);
ix = p1.x + a*(p2.x-p1.x);
iy = p1.y + a*(p2.y-p1.y);
iz = plane_z;
Edit3: added Orbling's solution which is slightly faster, and doesnt rely on two points necessarily.
You can implement a strait-forward solution like there http://paulbourke.net/geometry/planeline/, then apply your simplifications. In the algebraic solution (#2) A and B are zeros in your case (if i understand correctly this statement)
plane is always in the same z-axis (so it cannot be rotated)
Note: your line should be a point and a direction, or two points right?
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