判断四个点是否在同一个平面上

Posted 2021-07-15 momo尼

算出从同一个点出发的三个向量，如果这三个向量的x,y,z值所构成的行列式结果为零，则这四个点在同一平面上，否则不同一平面
如A,B,C，D四点
三个向量AB(x2-x1,y2-y1,z2-z1),AC(x3-x1,y3-y1,z3-z1),AD(x4-x1,y4-y1,z4-z1)(是向量，BA,AB是不同向量)
V=|[AB AC AD]|（混合积）
V= |x2-x1 y2-y1 z2-z1|
|x3-x1 y3-y1 z3-z1|
|x4-x1 y4-y1 z4-z1|
如果V=0，则四点共面
如果V!=0,则四点不同在一平面

#include<iostream>
#include<math.h>
#include<cmath>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
class Point
{
public:
	int x, y, z;
	Point(int a, int b, int c)
	{
		x = a;y = b;z = c;
	}
};
class isFlat
{
public:
	int x, y, z;
	isFlat(int a, int b, int c)
	{
		x = a;y = b;z = c;
	}
	int cul(isFlat& v2, isFlat& v3)
	{
		return(x * (v2.y) * (v3.z) + (v2.x) * (v3.y) * z + y * (v2.z) * (v3.x) - z * (v2.y) * (v3.x) - x * (v2.z) * (v3.y) - y * (v2.x) * (v3.z));
	}
};
int main()
{
	int n;
	cin >> n;
	while (n--)
	{
		int x1, y1, z1, x2, y2, z2, x3, y3, z3, x4, y4, z4;
		cin >> x1 >> y1 >> z1 >> x2 >> y2 >> z2 >> x3 >> y3 >> z3 >> x4 >> y4 >> z4;
		Point p1(x1, y1, z1), p2(x2, y2, z2), p3(x3, y3, z3), p4(x4, y4, z4);//记录四个点
		isFlat v1(x2 - x1, y2 - y1, z2 - z1), v2(x3 - x1, y3 - y1, z3 - z1), v3(x4 - x1, y4 - y1, z4 - z1);//由同一点出发的三个向量
		int result = v1.cul(v2, v3);//行列式结果
		if (result == 0)
		{
			cout << "Yes" << endl;
		}
		else
		{
			cout << "No" << endl;
		}
	}
	return 0;
}