已知平面内三点，求其平面的法向量

设三点坐标为A(x1,y1,z1),B(x2,y2,z2),C(x3,y3,z3)
向量AB=(x2-x1,y2-y1,z2-z1),AC=(x3-x1,y3-y1,z3-z1)
AB、AC所在平面的法向量即AB×AC=(a,b,c),其中：
a=(y2-y1)(z3-z1)-(z2-z1)(y3-y1)
b=(z2-z1)(x3-x1)-(z3-z1)(x2-x1)
c=(x2-x1)(y3-y1)-(x3-x1)(y2-y1)

设 a=(ax,ay,az)， b=(bx,by,bz)。i，j，k分别是X，Y，Z轴方向的单位向量，则：
a×b=(aybz-azby)i+(azbx-axbz)j+(axby-aybx)k
a·b=(axbx+ayby+az*bz)

Python代码

def normal_vector(p1, p2, p3):
    x1, y1, z1 = p1
    x2, y2, z2 = p2
    x3, y3, z3 = p3

    a = (y2 - y1) * (z3 - z1) - (z2 - z1) * (y3 - y1)
    b = (z2 - z1) * (x3 - x1) - (x2 - x1) * (z3 - z1)
    c = (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1)

    return (a, b, c)

if __name__ == '__main__':
    p1 = 1.0, 5.2, 0.0
    p2 = 2.8, 3.9, 1.0
    p3 = 7.6, 8.4, 2.0
    p4 = normal_vector(p1, p2, p3)
    print(p4)

    # output (-5.800000000000001, 3.0, 14.340000000000002)

C++代码文章来源地址https://uudwc.com/A/pjrk8

#include<iostream>

using namespace std;

//三维double矢量
struct Vec3d
{
	double x, y, z;

	Vec3d()
	{
		x = 0.0;
		y = 0.0;
		z = 0.0;
	}
	Vec3d(double dx, double dy, double dz)
	{
		x = dx;
		y = dy;
		z = dz;
	}
	void Set(double dx, double dy, double dz)
	{
		x = dx;
		y = dy;
		z = dz;
	}
};

//计算三点成面的法向量
void Cal_Normal_3D(const Vec3d& v1, const Vec3d& v2, const Vec3d& v3, Vec3d &vn)
{
	//v1(n1,n2,n3);
	//平面方程: na * (x – n1) + nb * (y – n2) + nc * (z – n3) = 0 ;
	double na = (v2.y - v1.y)*(v3.z - v1.z) - (v2.z - v1.z)*(v3.y - v1.y);
	double nb = (v2.z - v1.z)*(v3.x - v1.x) - (v2.x - v1.x)*(v3.z - v1.z);
	double nc = (v2.x - v1.x)*(v3.y - v1.y) - (v2.y - v1.y)*(v3.x - v1.x);

	//平面法向量
	vn.Set(na, nb, nc);
}

int main()
{	
	// Vec3d v1(1.0, 5.2, 3.7);
	// Vec3d v2(2.8, 3.9, 4.5);
	// Vec3d v3(7.6, 8.4, 6.2);   
	// 法向量为：-5.81 0.78 14.34;

	Vec3d v1(1.0, 5.2, 0.0);
	Vec3d v2(2.8, 3.9, 1.0);
	Vec3d v3(7.6, 8.4, 2.0);
	// 法向量为：-5.8 3 14.34
	
	Vec3d vn;
	Cal_Normal_3D(v1, v2, v3, vn);
	// cout <<"法向量为："<< vn.x << '\t' << vn.y << '\t' << vn.z << '\n';
    cout <<"法向量为："<< vn.x << " " << vn.y << " " << vn.z << '\n';

	return 0;
}