Getting two different intersection points for intersection of two lines in 3D

According to the link1 . If my input is like this

A1=[0 0 0];
A2=[5 5 5];

B1=[0 5 5];
B2=[5 0 0];

I will get the output same output points as points [2.5000 2.5000 2.5000]

However, if I change my input to the following.

    A1 =

       44.2709   -6.6805 -159.7093

    A2 =

        0.6680    4.4271         0

    B1 =

       48.4957   -8.7269 -160.0394

    B2 =

        0.8727    4.8496         0

I will get two different points of intersection as point1 =[1.2461 4.2798 -2.1173]

point1 =[1.4910 4.6733 -2.0778]

Does anyone know how to fix this issue?

----------------Code snapshot---------------------------------------------------------

    A1=[0 0 0];
     A2=[5 5 5];

    B1=[0 5 5];

    B2=[5 0 0];

    ts = [A2(:) - A1(:),-(B2(:)-B1(:))]\(B1(:) - A1(:))

     point1=A1 + (A2 - A1)*ts(1)

     point1=B1 + (B2 - B1)*ts(2)

Answer 1

[Matt J]

It means your lines do not intersect. However, the 2 solutions you're obtaining are the pair of points from line A and B of minimum possible distance from each other.

[Don Wei]

Thank you for your reply. However, I have check if these two lines are skewed or not through link2. It turns out that it is skew check is not equal to 0. This is why I am wondering what went wrong

[Matt J]

As the link mentions, if the "skew check" is not equal zero, it means the lines are skewed. In practice, it is impossible to have an ideal intersection of two lines anyway because of floating point precision limits. Your lines may "almost" intersect. The lines in this example are rather close together,

    >> point1=A1 + (A2 - A1)*ts(1);
    >> point2=B1 + (B2 - B1)*ts(2);
    >> norm(point1-point2)

    ans =

        2.3647