我想在路口的算法之间两架飞机

但都没有返回正确的结果。

but all not return correct results

我在3D 2矩形每个定义的三个点，我想在路口等线的两个点，这两个点的交汇点结束 我做以下步骤：

I have two rectangle in 3D each defined by three points , I want to get the two points on the line of intersection such that the two points at the end of the intersection I do the following steps:

我想获得线路的实际结束点所在的平面的边界

I want to get the actual end points of line that lie on the boundary of the plane

最好的问候

推荐答案

您提供最有可能的链接是否具有正确的解决方案:) 你有没有正确地改变你的三分信息的斧头+到+锆+ D = 0的形式？检查所有的点满足这个公式。如果你有正确的{A，B，C，D}，则很容易为链接描述来计算剩下的..

The link you provided most probably has the correct solution :) Did you correctly transform your three points info the Ax+By+Cz+D = 0 form? Check if all those points satisfy this formula. If you have the correct {A, B, C, D} then it's easy to calculate the rest as described in the link..

这里是说明如何开始使用3点这个公式的链接。

Here is a link which explains how to get this formula using 3 points.

好吧，这里一个简单的总结：

Ok, here a simple summary:

在空间鉴于三个点（X1，Y1，Z1），（X2，Y2，Z2），（X3，Y3，Z3），计算的：

Given three points in space (x1,y1,z1), (x2,y2,z2), (x3,y3,z3), calculate this:

A = Y1（Z2 - Z3）+ Y2（Z3 - Z1）+ Y3（Z1 - Z2）

A = y1 (z2 - z3) + y2 (z3 - z1) + y3 (z1 - z2)

B = Z1（X2 - X3）+ Z2（X3 - X1）+ Z3（X1 - X2）

B = z1 (x2 - x3) + z2 (x3 - x1) + z3 (x1 - x2)

C = X1（Y2 - Y3）+ X2（Y3 - Y1）+ X3（Y1 - Y2）

C = x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)

D = - （X1（Y2 Z3 - Y3 Z2）+ X2（Y3 Z1 - Y1 Z3）+ X3（Y1 Z2 - Y2 Z1））

D = -(x1 (y2 z3 - y3 z2) + x2 (y3 z1 - y1 z3) + x3 (y1 z2 - y2 z1))

有两个平面。这意味着你有A1，B1，C1，D1和A2，B2，C2，D2

for both planes. Which means you have A1, B1, C1, D1 and A2, B2, C2, D2.

使用A，B，C，D计算如下：

Using A, B, C, D calculate this:

X1 = 0

Z1 =（B2 / B1）* D1 - D2）/（C2 - C1 * B2 / B1）

z1 = (B2/B1)*D1 - D2)/(C2 - C1*B2/B1)

Y1 =（-C1 * Z1 - D1）/ B1

y1 = (-C1 * z1 - D1) / B1

那么这样的：

X2 =一定的价值。

x2 = some value..

Z2 =（B2 / B1）*（A 1 * 2 + D 1） - A 2 * X 2 - D 2）/（C 2 - C 1 * B2 / B1）

z2 = (B2/B1)*(A1 * x2 + D1) - A2 * x2 - D2)/(C2 - C1*B2/B1)

Y2 =（-C1 * Z2 -A1 * X2 - D1）/ B1

y2 = (-C1 * z2 -A1 * x2 - D1) / B1

基本上只是结合在这两个环节中描述这两种方法。

Basically just combine both ways described in those two links..