查找2D三角形在3D空间中是否垂直 [英] Find if 2 triangles are perpendicular in 3D space

问题描述

我在3D空间中有2个由3个点组成的三角形.

I have 2 triangles in 3D space made of 3 points.

我假设我需要使用点积，但是如何排列矩阵?

I assume I need to use the dot product but how do I arrange the matrix?

我想我有作品，但需要帮助来安排:)

I think I have the pieces but need help to arrange it :)

谢谢.

下面包含当前代码，不确信它是正确的.

Current code included below, not convinced it is correct.

vx1 = self.vertices[f[1]].x-self.vertices[f[0]].x
vy1 = self.vertices[f[1]].y-self.vertices[f[0]].y
vz1 = self.vertices[f[1]].z-self.vertices[f[0]].z

vx2 = self.vertices[f[2]].x-self.vertices[f[0]].x
vy2 = self.vertices[f[2]].y-self.vertices[f[0]].y
vz2 = self.vertices[f[2]].z-self.vertices[f[0]].z

plane1 = np.cross([vx1,vy1,vz1],[vx2, vy2, vz2])

vx3 = self.vertices[ff[1]].x-self.vertices[ff[0]].x
vy3 = self.vertices[ff[1]].y-self.vertices[ff[0]].y
vz3 = self.vertices[ff[1]].z-self.vertices[ff[0]].z

vx4 = self.vertices[ff[2]].x-self.vertices[ff[0]].x
vy4 = self.vertices[ff[2]].y-self.vertices[ff[0]].y
vz4 = self.vertices[ff[2]].z-self.vertices[ff[0]].z

plane2 = np.cross([vx3,vy3,vz3],[vx4, vy4, vz4])

print("p1",plane1)
print("p2",plane2)
print("dot",np.dot(plane1,plane2))

if np.dot(plane1,plane2) == 0:
    print("perpendictular")

推荐答案

@Rory Daulton的回答很直观，也很完美.我要补充一下，您可以使用 Binet-Cauchy公式十倍的加速:

@Rory Daulton's answer is intuitive and perfectly fine. I'd just add that you can use Binet-Cauchy formula for a ten-fold speedup:

import numpy as np

TR1, TR2 = np.random.random((2,3,3))

def xprod():
    return np.cross(*(TR1[:2]-TR1[2]))@np.cross(*(TR2[:2]-TR2[2]))

def bincau():
    aux = ((TR1[:2]-TR1[2])@(TR2[:2]-TR2[2]).T).ravel()
    return aux[0]*aux[3] - aux[1]*aux[2]

xprod()
# -0.04300263623056995
bincau()
# -0.043002636230569956

from timeit import timeit

timeit(xprod, number=100000)
# 7.751510428992333
timeit(bincau, number=100000)
# 0.620043026006897

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