math - 找到两条 3D 线的二等分

Question

我想计算两条具有相交点的 3D 线的二等分。这些线是由一个点和一个方向向量定义的共线线。我怎样才能找到它们二等分的两条线的方程？

标签： math3dgeometrysympy

Answer 1

A + t * dA, B + s * dB让线定义A, B为基点和dA, dB归一化方向向量。

如果保证线有交点，可以使用点积方法（改编自斜线最小距离算法）找到：

u = A - B
b = dot(dA, dB)
if abs(b) == 1: # better check with some tolerance
   lines are parallel
d = dot(dA, u)
e = dot(dB, u)
t_intersect = (b * e - d) / (1 - b * b)
P = A + t_intersect * dA

现在关于平分线：

bis1 = P + v * normalized(dA + dB)
bis2 = P + v * normalized(dA - dB)

快速检查 2D 案例

k = Sqrt(1/5) 
A = (3,1)     dA = (-k,2k)
B = (1,1)     dB = (k,2k)
u = (2,0)
b = -k^2+4k2 = 3k^2=3/5
d = -2k  e = 2k
t = (b * e - d) / (1 - b * b) = 
    (6/5*k+2*k) / (16/25) =  16/5*k * 25/16 = 5*k
Px = 3 - 5*k^2 = 2
Py = 1 + 10k^2 = 3
normalized(dA+dB=(0,4k)) =   (0,1)
normalized(dA-dB=(-2k,0)) = (-1,0)