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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?

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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).

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