Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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 $$ \lambda=-\frac{a'\cdot w'}{w'\cdot w'}\tag{2} $$

share|improve this answer
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

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.