Matlab 在3D中查找3条给定直线的旋转，直到与其他3条给定直线相交

我有一个包含3个方程和3个变量的matlab代码，我正试图解决（但没有成功）。这些方程是带有三个未知变量x、y、z的符号方程，其余符号是已知的。 我需要找到x，y，z的值作为其他已知符号的表达式。 以下是matlab代码：

clear;
close all;
clc;

%unknown variables
syms x y z %The angles of rotations (in radians)

%known symbols
syms px1 py1 pz1 %point on line1 set1
syms px2 py2 pz2 %point on line2 set1
syms px3 py3 pz3 %point on line3 set1

syms vx1 vy1 vz1 %vector of line1 set1
syms vx2 vy2 vz2 %vector of line2 set1
syms vx3 vy3 vz3 %vector of line3 set1   

syms qx1 qy1 qz1 %point on line1 set2
syms qx2 qy2 qz2 %point on line2 set2
syms qx3 qy3 qz3 %point on line3 set2

syms ux1 uy1 uz1 %vector of line1 set2
syms ux2 uy2 uz2 %vector of line2 set2
syms ux3 uy3 uz3 %vector of line3 set2

eq1 = qx1*(sin(x)*sin(z) - cos(x)*cos(z)*sin(y)) + qy1*(cos(z)*sin(x) + cos(x)*sin(y)*sin(z)) + ((ux1*(sin(x)*sin(z) - cos(x)*cos(z)*sin(y)) + uy1*(cos(z)*sin(x) + cos(x)*sin(y)*sin(z)) + uz1*cos(x)*cos(y))*(px1 - qz1*sin(y) - qx1*cos(y)*cos(z) + qy1*cos(y)*sin(z) + (vx1*(qx1*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) - py1 + qy1*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - qz1*cos(y)*sin(x) + ((ux1*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy1*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz1*cos(y)*sin(x))*(px1 - qz1*sin(y) - qx1*cos(y)*cos(z) + qy1*cos(y)*sin(z)))/(uz1*sin(y) + ux1*cos(y)*cos(z) - uy1*cos(y)*sin(z))))/(vy1 - (vx1*(ux1*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy1*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz1*cos(y)*sin(x)))/(uz1*sin(y) + ux1*cos(y)*cos(z) - uy1*cos(y)*sin(z)))))/(uz1*sin(y) + ux1*cos(y)*cos(z) - uy1*cos(y)*sin(z)) + qz1*cos(x)*cos(y) == pz1 + (vz1*(qx1*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) - py1 + qy1*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - qz1*cos(y)*sin(x) + ((ux1*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy1*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz1*cos(y)*sin(x))*(px1 - qz1*sin(y) - qx1*cos(y)*cos(z) + qy1*cos(y)*sin(z)))/(uz1*sin(y) + ux1*cos(y)*cos(z) - uy1*cos(y)*sin(z))))/(vy1 - (vx1*(ux1*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy1*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz1*cos(y)*sin(x)))/(uz1*sin(y) + ux1*cos(y)*cos(z) - uy1*cos(y)*sin(z)))
eq2 = qx2*(sin(x)*sin(z) - cos(x)*cos(z)*sin(y)) + qy2*(cos(z)*sin(x) + cos(x)*sin(y)*sin(z)) + ((ux2*(sin(x)*sin(z) - cos(x)*cos(z)*sin(y)) + uy2*(cos(z)*sin(x) + cos(x)*sin(y)*sin(z)) + uz2*cos(x)*cos(y))*(px2 - qz2*sin(y) - qx2*cos(y)*cos(z) + qy2*cos(y)*sin(z) + (vx2*(qx2*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) - py2 + qy2*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - qz2*cos(y)*sin(x) + ((ux2*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy2*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz2*cos(y)*sin(x))*(px2 - qz2*sin(y) - qx2*cos(y)*cos(z) + qy2*cos(y)*sin(z)))/(uz2*sin(y) + ux2*cos(y)*cos(z) - uy2*cos(y)*sin(z))))/(vy2 - (vx2*(ux2*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy2*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz2*cos(y)*sin(x)))/(uz2*sin(y) + ux2*cos(y)*cos(z) - uy2*cos(y)*sin(z)))))/(uz2*sin(y) + ux2*cos(y)*cos(z) - uy2*cos(y)*sin(z)) + qz2*cos(x)*cos(y) == pz2 + (vz2*(qx2*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) - py2 + qy2*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - qz2*cos(y)*sin(x) + ((ux2*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy2*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz2*cos(y)*sin(x))*(px2 - qz2*sin(y) - qx2*cos(y)*cos(z) + qy2*cos(y)*sin(z)))/(uz2*sin(y) + ux2*cos(y)*cos(z) - uy2*cos(y)*sin(z))))/(vy2 - (vx2*(ux2*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy2*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz2*cos(y)*sin(x)))/(uz2*sin(y) + ux2*cos(y)*cos(z) - uy2*cos(y)*sin(z)))
eq3 = qx3*(sin(x)*sin(z) - cos(x)*cos(z)*sin(y)) + qy3*(cos(z)*sin(x) + cos(x)*sin(y)*sin(z)) + ((ux3*(sin(x)*sin(z) - cos(x)*cos(z)*sin(y)) + uy3*(cos(z)*sin(x) + cos(x)*sin(y)*sin(z)) + uz3*cos(x)*cos(y))*(px3 - qz3*sin(y) - qx3*cos(y)*cos(z) + qy3*cos(y)*sin(z) + (vx3*(qx3*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) - py3 + qy3*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - qz3*cos(y)*sin(x) + ((ux3*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy3*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz3*cos(y)*sin(x))*(px3 - qz3*sin(y) - qx3*cos(y)*cos(z) + qy3*cos(y)*sin(z)))/(uz3*sin(y) + ux3*cos(y)*cos(z) - uy3*cos(y)*sin(z))))/(vy3 - (vx3*(ux3*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy3*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz3*cos(y)*sin(x)))/(uz3*sin(y) + ux3*cos(y)*cos(z) - uy3*cos(y)*sin(z)))))/(uz3*sin(y) + ux3*cos(y)*cos(z) - uy3*cos(y)*sin(z)) + qz3*cos(x)*cos(y) == pz3 + (vz3*(qx3*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) - py3 + qy3*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - qz3*cos(y)*sin(x) + ((ux3*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy3*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz3*cos(y)*sin(x))*(px3 - qz3*sin(y) - qx3*cos(y)*cos(z) + qy3*cos(y)*sin(z)))/(uz3*sin(y) + ux3*cos(y)*cos(z) - uy3*cos(y)*sin(z))))/(vy3 - (vx3*(ux3*(cos(x)*sin(z) + cos(z)*sin(x)*sin(y)) + uy3*(cos(x)*cos(z) - sin(x)*sin(y)*sin(z)) - uz3*cos(y)*sin(x)))/(uz3*sin(y) + ux3*cos(y)*cos(z) - uy3*cos(y)*sin(z)))


equations_system(1) = eq1;
equations_system(2) = eq2;
equations_system(3) = eq3;

fprintf('Trying to solve the equations (the function never finish)\n');
[x_solution, y_solution, z_solution] = solve(equations_system,[x, y, z]);


%The above equations are solving the following problem:
%Given 2 sets of 3 lines each (where each line is given by a point and a vector):
%set1={ [(px1,py1,pz1), (vx1,vy1,vz1)], [(px2,py2,pz2),(vx2,vy2,vz2)],[(px3,py3,pz3), (vx3,vy3,vz3)] } (the meaning of the symbols are written in their definition above).
%set2={ [(qx1,qy1,qz1), (ux1,uy1,uz1)], [(qx2,qy2,qz2),(ux2,uy2,uz2)],[(qx3,qy3,qz3), (ux3,uy3,uz3)] } (the meaning of the symbols are written in their definition above).
%find the angles of rotation x,y,z (in radians) such that:
%R=rotate_x*rotate_y*rotate_z
% rotate_x = [1 0 0; 0 cos(x) -sin(x) ; 0 sin(x) cos(x)];
% rotate_y = [cos(y) 0 sin(y); 0  1  0 ; -sin(y) 0 cos(y)];
% rotate_z = [cos(z) -sin(z) 0; sin(z) cos(z) 0 ; 0 0 1];
%and if we rotate set1 by R, then
%all the lines in set1 will
%intersect their corresponding lines in set2.
%meaning, let's define set1_r as set1 after the above rotation.
%then all the following will occure:
%   - the first line in set1_r intersects the first line in set2
%   - the second line in set1_r intersects the second line in set2
%   - the third line in set1_r intersects the third line in set2


%Here's an example of the two sets:
% px1=0;
% py1=0;
% pz1=-30;
% 
% px2=0;
% py2=0;
% pz2=-30;
% 
% px3=0; 
% py3=0;
% pz3=-30;
% 
% vx1 = -0.083717247687439; 
% vy1 = -0.107930827800543; 
% vz1 = 0.990627255252918;
% 
% vx2 = 0.076364294519742;
% vy2 = 0.060269029165473; 
% vz2 = 0.995256820446840;
% 
% vx3 = -0.081460429387834; 
% vy3 = 0.105021268850622; 
% vz3 = 0.991128009660183; 
% 
% qx1=0; 
% qy1=0;
% qz1=-30;
% 
% qx2=0; 
% qy2=0;
% qz2=-30;
% 
% qx3=0; 
% qy3=0; 
% qz3=-30;
% 
% ux1 = -0.079382581863774;
% uy1 = -0.095259098236529; 
% uz1 = 0.992282273297173;
% 
% ux2 = 0.079382581863774; 
% uy2 = 0.095259098236529; 
% uz2 = 0.992282273297173;
% 
% ux3 = -0.086165283952334; 
% uy3 = 0.103398340742801; 
% uz3 = 0.990900765451843;
% 

%example solution for the above values
%all zeros since all the lines pass through (0,0,-30)
% x=0;
% y=0;
% z=0;    

%another solution for the above values
% x=deg2rad(30); %x=0.523598775598299
% y=0;
% z=0;

%There are probably more solutions.

“solve”函数从未完成，因此我无法获得解决方案

有人能提出一个解方程的方法吗

matlab代码中的方程式解决了以下数学问题（在matlab中也以注释的形式编写，其中有一个数值示例，我在这里没有编写）：

在三维空间中给出2组3条线（每条线由一个点和一个向量给出）：

set1={[（px1，py1，pz1），（vx1，vy1，vz1）]，[（px2，py2，pz2），（vx2，vy2，vz2）]，[（px3，py3，pz3），（vx3，vy3，vz3）]

set2={[（qx1，qy1，qz1），（ux1，uy1，uz1）]，[（qx2，qy2，qz2），[（qx3，qy3，qz3）]

求旋转角度x、y、z（弧度），以便：

R=旋转x*旋转y*旋转z

旋转_x=[10；0 cos（x）-sin（x）；0 sin（x）cos（x）]

旋转y=[cos（y）0sin（y）；0110；-sin（y）0cos（y）]

旋转_z=[cos（z）-sin（z）0；sin（z）cos（z）0；0 0 1]

如果我们把set1旋转R，那么set1中的所有直线都会与set2中相应的直线相交

也就是说，让我们在上述旋转之后将set1_r定义为set1

然后set1中的线i（i=1,2,3）将与set2中的线i相交

如上所述，一个数值示例是用matlab中的注释编写的（包括我想要得到的期望角度，或者至少是其中的一部分，因为我不知道有多少解）。我没有在这里写数字示例，因为我的文章已经太长了

如果matlab中的方程无法求解（我希望不是…），有人能提出更好的解决数学问题的方法吗

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你应该把这个贴在数学上，用方程式而不是代码。你是对的。我刚刚在这个链接中发布了：这是否意味着matlab不能解方程？你应该在数学上发布这个，用方程而不是代码。你是对的。我刚刚在这个链接中发布了：这是否意味着matlab不能解方程？