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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

$$(a_1 + \bar{\lambda} w_1 - a_2) \cdot (w_2 \times (w_1 \times w_2)) = 0$$


$$\bar{\lambda} = \frac{(a_2 - a_1) \cdot (w_2 \times (w_1 \times w_2))}{\Vert w_1 \times w_2 \Vert^2}$$

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.

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up vote 1 down vote accepted

The line segment joining $(a+\lambda w)$ and $(a+\lambda w)+\epsilon(a'+\lambda w')$, should be perpendicular to both, so perpendicular to $(a+\lambda w)$ and $(a'+\lambda w')$. That means it would have direction $w\times w'$. Thus, the segment would lie in the plane $(a+\lambda w+\mu w\times w')$ and intersect $(a+\lambda w)+\epsilon(a'+\lambda w')$.

Any point $x$ in the plane $(a+\lambda w+\mu w\times w')$ satisfies $0=(x-a)\cdot(w\times(w\times w'))$, so we need to find the $\lambda$ where $$ \begin{align} 0&=((a+\lambda w)+\epsilon(a'+\lambda w')-a)\cdot(w\times(w\times w'))\\ &=\epsilon(a'+\lambda w')\cdot(w\times(w\times w')) \end{align} $$ Therefore, $$ \lambda=-\frac{a'\cdot(w\times(w\times w'))}{w'\cdot(w\times(w\times w'))}=\frac{a'\cdot(w\times(w\times w'))}{|w\times w'|^2}\tag{1} $$ If $|w|=1$, then $w'\perp w$, and therefore, $w\times(w\times w')=-w'$. Then $(1)$ simplifies to $$ \lambda=-\frac{a'\cdot w'}{w'\cdot w'}\tag{2} $$

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Thanks for your answer. I think I get it, but I am trying to make this a bit more rigorous (see my edit of the question above). I would be glad if maybe you could give me a hint at how to achieve this – koletenbert Sep 27 '11 at 18:36
Never mind, I think I figured it out :) – koletenbert Sep 27 '11 at 18:48

Consider the map $$\phi:(t,s)\mapsto a(t)+sw(t).$$ The determinant of its Jacobian is $\det(a'(t),w(t))+s\det(w'(t),w(t))$. This vanishes exactly when $$s=-\frac{\det(a'(t),w(t))}{\det(w'(t),w(t))},$$ so we get a curve of bad points $$\beta:t\mapsto a(t)-\frac{\det(a'(t),w(t))}{\det(w'(t),w(t))}w(t).$$ This curve $\beta$ is tangent to each of your lines (in other words, it is an envelope to the family) I think it is the locus of your limit points.

For example, if $w(t)=n(t)$ is the normal vector to the curve, then the curve $\beta$ is the evolute to the original curve $a$, the locus of its centers of curvature.

A random example: the curve $a(t)=(t, t^2/4 + \cos t)$ and $w(t)$ the normalization of $(t^3 + 1, 2 t - 3)$. Here is a picture of the curve in blue, the lines in gray, and the envelope in red:

enter image description here

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Thanks for your answer... I realize that I left out an important detail in the original problem: the lines are in 3D (hence the line segment joining the skew lines). Sorry for the confusion, I will edit accordingly. – koletenbert Sep 26 '11 at 20:25

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