固定 z 轴上的简单线平面相交？
如果我知道平面始终位于同一 z 轴（因此无法旋转），并且其宽度/高度是无限的，那么有什么快速方法可以检查我的线在平面中的何处相交？另外，我的“线”实际上不是一条线，而是一个 3d 矢量，因此“线”可以延伸到无限远的距离。
下面是依赖两点的代码： （p1 和 p2 是直线的起点和终点。plane_z = 平面所在的位置）
k1 = -p2.z/(p1.z-p2.z-plane_z);
k2 = 1.0f-k1;
ix = k1*p1.x + k2*p2.x;
iy = k1*p1.y + k2*p2.y;
iz = plane_z; // where my plane lays
另一个使用向量的解决方案（我也像第一个示例一样使用两个点，“p2.x-p1.x”等是向量计算）：
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：添加了Orbling的解决方案，该解决方案稍快一些，并且不一定依赖于两个点。
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;
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.
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您可以实现像 http://paulbourke.net/geometry/planeline/ 这样的直接解决方案，然后应用您的简化。在代数解（＃2）中，A和B在你的情况下为零（如果我正确理解了这个陈述）
注意：您的线应该是一个点和一个方向，或者是两个点，对吗？
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)
Note: your line should be a point and a direction, or two points right?