Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

During tutoring (12th grade, regular Math class), I had to explain how to find the two points $s$ and $s'$ that are the base of the perpendicular connection between two skew straight lines $g$ and $h$.

Where $g$ and $h$ are given respectively as: $$ g \colon \mathbb R \to \mathbb R^3; t \mapsto \vec x = \vec x_0 + \vec v_g t $$

What I did is to calculate a normal vector to both $\left(\vec n \propto \vec v_g \times \vec v_h \right)$. Then move $h$ along $\vec n$ so that $h'$ intersects $g$. Then I would just calculate the intersection $s$ of $g$ and $h'$, and move the intersection back down to $h$, yielding $s'$.

Is there any shorter way?

share|cite|improve this question
up vote 1 down vote accepted

The difference $$\vec x_g + \vec v_g s - \vec x_h - \vec v_h s'$$ must be orthogonal to both $\vec v_g$ and $\vec v_h$. This gives a pair of linear equations for $s$ and $s'$. Moreover, the determinant of the system is $\|\vec v_g\|^2 \|\vec v_h\|^2 - (\vec v_g\cdot \vec v_h)^2$, which is $0$ iff the two lines are parallel (or coincident).

share|cite|improve this answer

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.