3D线 - 平面相交？ [英] 3D Line - Plane intersection?

问题描述

我有两个向量（X，Y，Z），一个高于 Y = 0 ，另一个低于 Y = 0 。
我想找到两个原始矢量之间的线与 Y = 0 级别相交的矢量（X，Y，Z）。
我该怎么做？

示例点A：

  X = -43.54235 
 Y = 95.2679138 
 Z = -98.2120361 
  

示例点B：

  X = -43.54235 
 Y = 97.23531 
 Z = -96.24464 
  

这些点从用户单击的两个UnProjections中读取，我试图将未投影到 Y = 0 。

（我发现 3D线平面相交，简单的平面，但不理解接受的答案，因为它是2D）

解决方案

我猜想通过两个向量，你真的意味着两个点，并且想要将连接这两个点的线与由 Y = 0 。

如果是这种情况，那么您可以使用两点之间的一条线的定义：

a +（D-A）* u，B +（E-B）* u，C +（F-C）* u> p>

其中< A，B，C> 是您的要点之一，< D ，E，F> 是另一点。 u 是一个未定义的标量，用于计算沿此线的点。

由于您正在相交与平面 Y = 0 一致，你只需要找到Y段为0的那一行上的点。

具体来说，解决 B +（E - B）* u = 0 中的 u ，然后将其反馈回原始线方程以找到X和Z分量。

I am having two Vectors (X,Y,Z), one above Y=0 and one below Y=0. I want to find the Vector (X,Y,Z) where the line between the two original vectors intersects with the Y=0 level. How do I do that?

Example Point A:

X = -43.54235
Y = 95.2679138
Z = -98.2120361

Example Point B:

X = -43.54235
Y = 97.23531
Z = -96.24464

These points read from two UnProjections from a users click and I'm trying to target the unprojection to Y=0.

(I found 3D line plane intersection, with simple plane but didn't understand the accepted answer as it's for 2D)

解决方案

I suspect that by two vectors, you really mean two points, and want to intersect the line connecting those two points with the plane defined by Y=0.

If that's the case, then you could use the definition of a line between two points:

<A + (D - A)*u, B + (E - B)*u, C + (F - C)*u>

Where <A,B,C> is one of your points and <D,E,F> is the other point. u is an undefined scalar that is used to calculate the points along this line.

Since you're intersecting this line with the plane Y=0, you simply need to find the point on the line where the "Y" segment is 0.

Specifically, solve for u in B + (E - B)*u = 0, and then feed that back into the original line equation to find the X and Z components.

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