Is there only one point where the line connecting the two skew lines is perpendicular to them both? And is that point always always always the place of closest approach?
There is a standard proof of this written down someplace but I forget it.
You start out by finding the closest point Q on a line L1 to a specified point P.
You should notice that:
(I) Then line-segment QP is perpendicular to L1. Do you see why this must always be the case?
(II) If Q were not the closest point on L1 to P, then QP would not be perpendicular to L1.
Now consider that P is on another line L2 ...
1. If L2 is parallel to L1, then |PQ| is the distance of closest approach and PQ is perpendicular to both.
2. If L2 intersects L1, then |PQ| is not the distance of closest approach unless |PQ|=0.
3. If L2 is skew to L1 ... then we already have QP perpendicular to L1 beacuse Q is the closest point on L1 to P. But P is not the closest point to Q on L2 unless PQ is perpendicular to L2 ...
because of statement (II) above.
i.e.
For |PQ| to be the distance of closest approach for the two skew lines, P must be the closest point on L2 to Q and Q must be the closest point on L1 to P ... so PQ must be perpendicular to both lines.
Since L1 and L2 are lines there can be only one (P,Q) pair such that |PQ| is the distance of closest approach - i.e. the solution is unique.