Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 68 / 1 / Solution

Parameterize the two line segments byPut $w=A-C$ and abbreviate $a=u\cdot u$, $b=u\cdot v$, $c=v\cdot v$, $d=u\cdot w$, and $e=v\cdot w$. Minimizing the Euclidean distance is equivalent to minimizing the convex quadratic functionover a closed square. For nonzero nonparallel segment directions, $\Delta =ac-b^2=\Vert u\times v{\Vert }^2>0$. An interior minimum must satisfywhose unique solution isKeep it as a candidate only if both parameters lie in $[0,1]$.

Every remaining minimum lies on the boundary. Define $[x{]}_{[0,1]}=\max (0,\min (1,x))$. The four edge minima are the point-to-segment orthogonal projectionsEvaluate $F$ at those four pairs and at the admissible interior pair, and return a pair with the smallest value. This solves closest points on two line segments completely: continuity guarantees a minimum, and every interior or boundary location has been tested. Do not independently clamp $s_0$ and $t_0$: the mixed term couples them. For example, $A=(0,0)$, $B=(1,0)$, $C=(2,-2)$ and $D=(3,-1)$ give the line minimizer $(s_0,t_0)=(4,2)$. Its independently clamped pair $(1,1)$ has squared distance $5$, whereas the correct edge minimizer $(1,1/2)$ has squared distance $9/2$.

Without general position, explicitly handle these configurations:

In floating-point arithmetic, almost-parallel directions require a condition-aware or higher-precision solve; they must not be declared exactly parallel solely because the computed determinant is small. The returned distance is $\boxed{\sqrt{\min F}}$, and the returned points are obtained by substitution into $P(s),Q(t)$.