# Is this lemma about the minimal distance of two lines true?

In school, I recently proved a solid geometry excercise by assuming that the following lemma is true:

If two lines $g$ and $h$ in the euclidian space are not parallel, and if the lines seem parallel under a parallel projection, then the distance between projected lines is equal to the minimal distance of the $g$ and $h$.

Is this true, and if yes, how can I prove it? Please try to prove it in a way, that is understandable for somebody in highschool. (An answer of the style "This is just a special case of lemma X" is not very helpful)

-
I don't understand this question. In Euclidean space, if two lines are not parallel, then they intersect (as lines are infinitely long) so the minimum distnace of $g$ and $h$ is necessarily $0$. –  Alex Becker Apr 28 '11 at 18:02
@Alex: You seem to be thinking of the Euclidean plane; in three-dimensional Euclidean space, skew lines are neither parallel, nor do they intersect. –  joriki Apr 28 '11 at 18:14
@Joriki: Boy do I feel stupid. I just figured that he was talking about plane geometry because that's all we talked about in HS. This question makes a lot more sense now. –  Alex Becker Apr 28 '11 at 21:46

If the lines appear parallel under a parallel projection, they must lie in two parallel planes perpendicular to the projection plane. Imagine these two planes superimposed. Then the square of the distance between two points on the lines is by Pythagoras given by the sum of the square of the distance within the planes and the square of the distance between the planes. If the lines are not parallel, they must intersect in the superimposed planes, so there must be a point where that squared distance takes its minimal value, the square of the distance between the planes. Thus the minimal distance is the distance between the planes, which is the distance between the projected lines.

-

Seems right.

Let G be point on g and H point on h such that GH is minimal distance between g and h.

Then your lemma follows from the fact that projection plane will be necessarily parallel to GH.

This is not a fool proof, sorry, but a part of it anyway.

-