根据两坐标系中公共点计算转换罗德里格斯参数，并以此参数进行空间直角坐标转换，c#，带详细注释

好的，这里是代码实现，注释已经尽量详细了：

using System;
using System.Numerics;

public class Program
{
    public static void Main(string[] args)
    {
        // 定义两个坐标系的公共点
        Vector3 p1 = new Vector3(1, 0, 0);
        Vector3 p2 = new Vector3(0, 1, 0);
        Vector3 p3 = new Vector3(0, 0, 1);

        Vector3 p1_new = new Vector3(0, 0, 0);
        Vector3 p2_new = new Vector3(0, 0, 1);
        Vector3 p3_new = new Vector3(-1, 0, 0);

        // 计算两个坐标系的基向量
        Vector3 x1 = Vector3.Normalize(p2 - p1);
        Vector3 y1 = Vector3.Normalize(p3 - p1);
        Vector3 z1 = Vector3.Normalize(Vector3.Cross(x1, y1));

        Vector3 x2 = Vector3.Normalize(p2_new - p1_new);
        Vector3 y2 = Vector3.Normalize(p3_new - p1_new);
        Vector3 z2 = Vector3.Normalize(Vector3.Cross(x2, y2));

        // 计算旋转矩阵
        Matrix4x4 R = Matrix4x4.Identity;
        for (int i = 0; i < 3; i++)
        {
            for (int j = 0; j < 3; j++)
            {
                // 计算旋转矩阵的每个元素
                float cos = Vector3.Dot(new Vector3(x1[i], y1[i], z1[i]), new Vector3(x2[j], y2[j], z2[j]));
                Vector3 n = Vector3.Cross(new Vector3(x1[i], y1[i], z1[i]), new Vector3(x2[j], y2[j], z2[j]));
                float sin = n.Length();
                n = Vector3.Normalize(n);

                Matrix4x4 R_temp = Matrix4x4.CreateFromAxisAngle(n, (float)Math.Atan2(sin, cos));

                // 矩阵相乘，得到最终的旋转矩阵
                R = R_temp * R;
            }
        }

        // 进行坐标转换
        Vector3 p = new Vector3(1, 1, 1);
        Vector3 p_new = Vector3.Transform(p - p1, R) + p1_new;

        Console.WriteLine(p_new);
    }
}

首先定义两个坐标系的公共点p1、p2、p3和p1_new、p2_new、p3_new，然后分别计算出它们的基向量x、y、z和x_new、y_new、z_new。接着通过循环计算旋转矩阵的每个元素，具体方法是：将两个基向量的i、j分量组成向量，计算其点积和叉积，以及旋转角度，然后使用CreateFromAxisAngle方法生成旋转矩阵R_temp，最后将R_temp与R相乘得到最终的旋转矩阵R。最后将需要转换的点p进行变换，得到在新坐标系下的坐标p_new。