Prove that there exists a shortest distance between two skew lines 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.

It can be done. We have
$$\frac{\partial R}{\partial s} (s,t)= 2(\mathbf{p}-\mathbf{q}+s\mathbf{u}+t\mathbf{v})\cdot \mathbf{u}, \quad \frac{\partial R}{\partial t} (s,t)= 2(\mathbf{p}-\mathbf{q}+s\mathbf{u}+t\mathbf{v})\cdot \mathbf{v}.$$
The condition for local extremum is
$$\frac{\partial R}{\partial s} (s_0,t_0) = \frac{\partial R}{\partial t} (s_0,t_0)=0$$
so such $s_0,t_0$ satisfy $\mathbf{p}-\mathbf{q}+s_0\mathbf{u}+t_0\mathbf{v} \perp \mathbf{u},\mathbf{v}$. Therefore, there is a scalar $\alpha$ such that
$$\mathbf{p}-\mathbf{q}+s_0\mathbf{u}+t_0\mathbf{v} = \alpha(\mathbf{u} \times \mathbf{v}).$$
Note that $\mathbf{u}$ and $\mathbf{v}$ are linearly independent (since the lines are skew) so $\{\mathbf{u}, \mathbf{v}, \mathbf{u} \times \mathbf{v}\}$ is a basis for $\Bbb{R}^3$ and hence $s_0, t_0, \alpha$ exist and are unique. Hence, if we know that this is a local minimum (e.g. by calculating the Hessian), it has to be a global minimum.
Scalar multiplying the above relation by $\mathbf{u} \times \mathbf{v}$, we get
$$(\mathbf{p}-\mathbf{q})\cdot (\mathbf{u} \times \mathbf{v}) = (\mathbf{p}-\mathbf{q}+s_0\mathbf{u}+t_0\mathbf{v}) \cdot (\mathbf{u} \times \mathbf{v})= \alpha \|\mathbf{u} \times \mathbf{v}\|^2$$
so $$\alpha = \frac{(\mathbf{p}-\mathbf{q})\cdot (\mathbf{u} \times \mathbf{v})}{\|\mathbf{u} \times \mathbf{v}\|^2}.$$
The minimal distance is now given by
$$\|\mathbf{p}-\mathbf{q}+s_0\mathbf{u}+t_0\mathbf{v}\| = \alpha \|\mathbf{u} \times \mathbf{v}\| = \frac{(\mathbf{p}-\mathbf{q})\cdot (\mathbf{u} \times \mathbf{v})}{\|\mathbf{u} \times \mathbf{v}\|}.$$
