Given is a one-parameter family of lines,
$L(t) = \{ a(t) + \lambda w(t) : \quad \lambda \in \mathbb{R} \}$
in which the base point $a$ and the direction vector $w$ vary smoothly with a parameter $t$ (you may assume that $|w| = 1$ and $w' \neq 0$).
There is a unique line segment that joins two nearby lines $L(t)$ and $L(t + \Delta t)$ orthogonally. Let $C$ be the center of this line segment.
My question: what is the position of $C$ in the limit, when $\Delta t \rightarrow 0$?
EDIT: I realize I should have mentioned that the lines are in $\mathbb{R}^3$, and therefore $L(t)$ and $L(t + \Delta t)$ are supposed to be skew lines. Also, the solution $C$ is known to be located at
\bar{\lambda} = -\frac{a'(t) \cdot w'(t)}{w'(t) \cdot w'(t)}
for a given $t$. I just don't know how to prove this.
EDIT 2: Based on robjohn's answer below, I've got the following:
Let $t_1$ and $t_2 = t_1 + \Delta t$ be two nearby values of the parameter $t$, corresponding to two lines $L_1$ and $L_2$.
The perpendicular that joins $L_1$ and $L_2$ has the direction $w_1 \times w_2$, and the plane containing $L_2$ and the perpendicular therefore has the normal
$w_2 \times (w_1 \times w_2)$
which means that, for any point $x$ in the plane, we get
$(x - a_2) \cdot (w_2 \times (w_1 \times w_2)) = 0$
We need to find the point $a_1 + \bar{\lambda} w_1$ at which $L_1$ intersects this plane, therefore
$(a_1 + \bar{\lambda} w_1 - a_2) \cdot (w_2 \times (w_1 \times w_2)) = 0$
or
$\bar{\lambda} = \frac{(a_2 - a_1) \cdot (w_2 \times (w_1 \times w_2))}{\Vert w_1 \times w_2 \Vert^2}$
This yields $\bar{\lambda}$ for any finite displacement $\Delta t = t_2 - t_1$. For $\Delta t \rightarrow 0$, the expression gets indeterminate (because $w_2 \rightarrow w_1$ implies $w_1 \times w_2 \rightarrow 0$)...
I believe I am close, but I don't see how to obtain the correct limit from this.