CGAL users discussion list

["Vu, Khuong" <>]

Step 1 is not clear. My mistake. Please correct it as below:

Find the plan (p') containing d1 and parallel to d2. Find the plan (p) 
containing d1 and perpendicular to (p').
The following steps are as presented earlier. 

In the special case when d1//d2, find the plan perpendicular to both lines. 
The line connecting 2 intersection is as required.

Best,
Khuong
________________________________________
From: Vu, Khuong
Sent: Friday, June 11, 2010 10:27 AM
To: 

Subject: RE: [cgal-discuss] shortest line between two lines in 3d

I don't know much about cgal for 3D but we can have a solution for your 
question.

Let 2 lines be d1 and d2.
1. find the plane (p) that contains the d1 and perpendicular to d2
2. the intersection of d2 with p is p2.
3. find the intersection (p1)of the line passing p1 and perpendicular to d1.

It can be proved that p1p2 is the required line (segment).

Best,
Khuong
________________________________________
From: clemen 
[]
Sent: Friday, June 11, 2010 10:13 AM
To: 

Subject: [cgal-discuss] shortest line between two lines in 3d

simple problem: i have two lines/rays in 3d and want to get the shortest line
that connects both (or similarly: the 3d point that has the shortest
distance to both lines).

i can only find the intersect method but this fits just for a special case.
Anybody knows if cgal can help here?
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