查找两条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。线是由点和方向向量定义的交线。如何找到两等分的两条线的方程式？

Answer 1

将直线定义为A + t * dA, B + s * dB，其中A, B是基点，dA, dB是归一化方向矢量。

如果可以保证线有交点，则可以使用点积方法（根据斜线最小距离算法进行调整：：]

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)

快速检查二维情况

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)

Answer 2

Python实现：

from sympy.geometry import Line3D, Point3D, intersection


# Normalize direction vectors:
def normalize(vector: list):
    length = (vector[0]**2 + vector[1]**2 + vector[2]**2)**0.5
    vector = [i/length for i in vector]
    return vector

# Example points for creating two lines which intersect at A
A = Point3D(1, 1, 1)
B = Point3D(0, 2, 1)

l1 = Line3D(A, direction_ratio=[1, 0, 0])
l2 = Line3D(A, B)

d1 = normalize(l1.direction_ratio)
d2 = normalize(l2.direction_ratio)

p = intersection(l1, l2)[0]  # Point3D of intersection between the two lines

bis1 = Line3D(p, direction_ratio=[d1[i]+d2[i] for i in range(3)])
bis2 = Line3D(p, direction_ratio=[d1[i]-d2[i] for i in range(3)])