Styjun

Running the code below results in two sets of solutions being given. For one set of solutions, the z-values will both be >=0 (39.4962 and 0). For the other set of solutions, one z-value will be <0 (-39.4962 and 0).

If I include the "assumeAlso" lines that are currently commented out, no solutions are given. This is not what I expected; I assumed only the second set of solutions would be disqualified, since the first solution does not contradict the assumptions.

Can anyone enlighten me as to why this is happening? Thanks in advance.

points = [0.368,0.02,2.3 ; -0.536,-0.108,2.3];
d = 40;

syms x1 y1 z1 x2 y2 z2 real


% assumeAlso(z1 >= 0)
% assumeAlso(z2 >= 0)


% Conditions L1 must satisfy
line1 = [
 x1/points(1,1) == y1/points(1,2)
 y1/points(1,2) == z1/points(1,3)
];

% Conditions L2 must satisfy
line2 = [
 x2/points(2,1) == y2/points(2,2)
 y2/points(2,2) == z2/points(2,3)
];


distance = [
 ( (x1-x2).^2 + (y1-y2).^2 + (z1-z2).^2 ) == d.^2
];

solved = solve([line1,line2,distance],[x1,y1,z1,x2,y2,z2]);


disp([ 
eval([solved.x1 solved.y1 solved.z1])
eval([solved.x2 solved.y2 solved.z2])
])

You are trying to solve a problem that has infinite solutions. In other words, there is an independency between the variables x1 y1 z1 x2 y2 z2 you are solving for. The solver seems to find this independency only when constraining the variable (and confusingly so only outputs a finite number of solutions if you don't): you will find that you get the same warning even when specifying something silly like

assumeAlso(z1>=-inf)
assumeAlso(z2>=-inf)

An inequality constraint like assumeAlso(z1>=0) does not remove the independency. Instead, imposing an equality constraint like assumeAlso(z2==0) will solve the problem. You can then specify assumeAlso(z1>=0) to find the one specific solution you are looking for. I.e., specify:

assumeAlso(z1>=0)
assumeAlso(z2==0)

However, note that, for example, you will find another feasible solution for the constraint assumeAlso(z2==1), or even assumeAlso(z2==2*z1), etc...