如何使用平面和线的顶点确定线段/射线是否与 3D 空间中的平面相交

Question

我有平面的四个角顶点以及线段的起点和终点的顶点。我希望能够检查该线是否与 3-D 空间中的平面相交。

我最初使用这个 https://www.youtube.com/watch?v=_W3aVWsMp14 视频来进行数学计算，将其转换为参数，然后求解。

但是，我对如何实际求解和格式化方程有点迷失。

import numpy as np
from sympy import Point3D, Plane, Line3D
import sympy as sp


def find_intersection_point(line_vertices, plane_vertices):

    t, tt = sp.symbols('t tt')

    line_start = np.array(line_vertices[0])
    line_end = np.array(line_vertices[1])
    line_direction = line_end - line_start

    line_x = line_start[0]*t + line_direction[0]
    line_y = line_start[1]*t + line_direction[1]
    line_z = line_start[2]*t + line_direction[2]

    print(line_x)
    print(line_y)
    print(line_z)

    # Find two vectors in the plane
    vector1 = np.subtract(plane_vertices[1], plane_vertices[0])
    vector2 = np.subtract(plane_vertices[2], plane_vertices[0])

    # Find the normal vector
    normal_vector = np.cross(vector1, vector2)
    print("normal vector: ", normal_vector)

    # normal_vector[0]*(line_x-plane_vertices)



def get_line_vetices(line):
    line_vertices = [[0 for i in range(3)] for j in range(2)]
    line_vertices = np.empty((2, 3))
    for i in range(2):
        line_vertices[i][0] = line.Shape.Vertexes[i].X
        line_vertices[i][1] = line.Shape.Vertexes[i].Y
        line_vertices[i][2] = line.Shape.Vertexes[i].Z
    return line_vertices


def get_plane_vertices(plane):
    plane_vertices = [[0 for i in range(3)] for j in range(4)]
    plane_vertices = np.empty((4, 3))
    for i in range(4):
        plane_vertices[i][0] = plane.Shape.Vertexes[i].X
        plane_vertices[i][1] = plane.Shape.Vertexes[i].Y
        plane_vertices[i][2] = plane.Shape.Vertexes[i].Z
    return plane_vertices


# Define the vertices of the plane as [x, y, z]
plane_vertices = [
    [-26.551074, 29.194627000000004, 0.0],
    [45.091893, 29.194627000000004, 0.0],
    [45.091893, -34.722119000000006, 0.0],
    [-26.551074, -34.722119000000006, 0.0]
]

plane_vertices = np.array(plane_vertices)

# Define the start and end points of the line segment
line_vertices = [
    [-112.944015, -52.98405, 0.0],
    [136.40151699999996, 39.730347999999985, 0.0]
]

line_vertices = np.array(line_vertices)

find_intersection_point(line_vertices, plane_vertices)

下面的代码是我正在尝试但最终没有成功的另一种方法。

# plane Points
a1 = Point3D(-26.551074, 29.194627000000004, 0.0)
a2 = Point3D(45.091893, 29.194627000000004, 0.0)
a3 = Point3D(45.091893, -34.722119000000006, 0.0)
# line Points
p0 = Point3D(-112.944015, -52.98405, 0.0)  # point in line
line_direction = np.subtract(line_vertices[0], line_vertices[1])

print(line_direction)

# create plane and line
plane = Plane(a1, a2, a3)

line = Line3D(p0, direction_ratio=line_direction)


print(f"plane equation: {plane.equation()}")
print(f"line equation: {line.equation()}")

# find intersection:
# Find intersection
intr = plane.intersection(line)
if intr:
    intersection = intr[0]
    print("It intersects")
    print(f"Intersection: {intersection}")
else:
    print("No intersection found.")

Answer 1

有了三个平面点

a1,a2,a3

和线点

p0, p1

，您可以使用向量来制作参数方程

 a12 = a2 - a1
 a13 = a3 - a1
 p01 = p1 - p0

飞机：

 x = a1.x + u * a12.x  + v * a13.x
 y = a1.y + u * a12.y  + v * a13.y
 z = a1.z + u * a12.z  + v * a13.z

线路：

 x = p0.x + t * p01.x
 y = p0.y + t * p01.y
 z = p0.z + t * p01.z

现在将直线方程代入平面方程：

 a1.x + u * a12.x  + v * a13.x - p0.x - t * p01.x = 0
 a1.y + u * a12.y  + v * a13.y - p0.y - t * p01.y = 0
 a1.z + u * a12.z  + v * a13.z - p0.z - t * p01.z = 0

并求解该线性方程组的未知数

u, v, t

如果系统判别式为零，则这条线位于平面内或平行于平面。

否则你就有了交点的参数并且可以计算它。

如果您有射线，请检查该参数

t >= 0

如果您有线段，请检查

t

参数是否在

0..1

范围内 如果您需要飞机的某些部分 - 检查

u,v

参数。对于由

a1,a2,a3

点（和隐式四个顶点）定义的矩形（更一般 - 平行四边形）来说，参数应在

0..1

范围内