将第二个方程代入第一个方程中,得到:
x0*(x1*x0/x+y1*y0/y+z1*z0/z-x1)+y0*(y1*y0/y+x1*x0/x+z1*z0/z-y1)+z0*(z1*z0/z+x1*x0/x+y1*y0/y-z1) = 0
整理得到:
(x1*x0/x+y1*y0/y+z1*z0/z-x1)*x0 + (y1*y0/y+x1*x0/x+z1*z0/z-y1)*y0 + (z1*z0/z+x1*x0/x+y1*y0/y-z1)*z0 = 0
这是一个三元一次方程组,可以使用高斯消元法或矩阵求逆的方法求解。下面是一个示例代码,用于解决这个方程组:
```c++
#include
#include
using namespace cv;
bool solveEquation(const Point3d& p0, const Point3d& p1, Point3d& solution)
{
double A[3][3] = {
{p1.x * p0.x / p.x + p1.y * p0.y / p.y + p1.z * p0.z / p.z - p1.x, p1.y * p0.y / p.y + p1.x * p0.x / p.x + p1.z * p0.z / p.z - p1.y, p1.z * p0.z / p.z + p1.x * p0.x / p.x + p1.y * p0.y / p.y - p1.z},
{p0.x, p0.y, p0.z},
{1 / p0.x, 1 / p0.y, 1 / p0.z}
};
double b[3] = {-p0.x, -p0.y, -1 / p0.z};
Mat A_mat(3, 3, CV_64FC1, A);
Mat b_mat(3, 1, CV_64FC1, b);
Mat x_mat;
bool success = solve(A_mat, b_mat, x_mat, DECOMP_LU);