TITLE: Prove that there exists a shortest distance between two skew lines
QUESTION [6 upvotes]: In every geometry textbook it states that the shortest distance between two skew lines (lines that are not co-planar) is given by the unique line that runs perpendicular to both skew lines. This is fairly simple to prove given a shortest line exists (see my comments below). However, how can we prove that between any two skew lines there exists a shortest straight path?
I have tried using calculus to show that for lines $L_1 = \mathbf{p}+s\mathbf{u}$ and $L_2 = \mathbf{q} + t\mathbf{v}$ the equation:
$$R(s,t) = \Vert \mathbf{p}+s\mathbf{u}-(\mathbf{q}+t\mathbf{v})\Vert^2$$

Has a local minimum

However I am not quite able to show that (without getting into page and pages of calculations) and what is more, even after I have shown this, it only proves a local minimum exists.

REPLY [6 votes]: If you have two skew lines $a$ and $b$ it is easy to construct a line perpendicular to both, hence proving that it exists.

Construct the plane $\beta$ containing $b$ and parallel to $a$.

Construct the plane $\alpha$ containing $a$ and perpendicular to $\beta$.

If $B$ is the intersection of $\alpha$ with $b$, then the line $AB$ passing through $B$ and perpendicular to $a$ is also perpendicular to $b$ and is thus the solution.



It is then immediate to show that $AB$ is the line of minimum distance: given any two points $P\in a$ and $Q\in b$, if $H$ is the projection of $P$ on $\beta$ we have:
$$
PQ^2=PH^2+HQ^2\ge PH^2=AB^2.
$$