Before you can read the shortest-distance formula, you must be able to read every mark inside it. We will build them one at a time, each on top of the last. Nothing is used before it is drawn.
Why do we need this? Because a line has to start somewhere. In the parent topic, a1 and a2 are just "a known point I picked on each wire."
Figure s01 shows the arrow a=(2,−1,3): the burnt-orange arrow runs from the origin O to the teal point, and the three dotted helper legs spell out the walk "2 East, then 1 South (that is −1 North), then 3 Up." Read the arrow as "the recipe to reach that point."
The little hat, as in i^,j^,k^, means a unit arrow (length exactly 1) pointing along the East, North, and Up axes. So (2,−1,3) and 2i^−j^+3k^ say the same thing — one uses commas, the other names each axis explicitly.
A position vector answers where; a direction vector answers which way. The parent uses b1,b2 for the two wires' directions. See Equation of a Line in 3D for how a point plus a direction is a whole line.
Figure s02 draws this: the orange point is a (the setting λ=0), the small orange arrow is one step of b, and the plum and dark dots mark λ=1 and λ=2. Notice every dot lands on the same teal line — turning the λ dial just slides you along it.
Why the square root of squares? It is the Pythagoras theorem in 3D: the arrow is the long diagonal of a box with sides vx,vy,vz. The final shortest distance is a length, so it must come out of a ∣⋯∣.
Why does the topic need it? To measure how much of the gap vector points in a chosen direction — that projected amount will become the shortest distance once the chosen direction is the both-perpendicular one. The key fact:
Figure s03 shows a plum arrow u and a teal arrow v. Drop a dotted vertical line from the tip of u onto v: the orange arrow is u's shadow on v. The dot product multiplies that shadow's length by ∣v∣ — so it literally measures "how much of u lies along v."
Here is why picking any points a1,a2 is safe: sliding along a wire adds a multiple of b to the gap, and (once we meet n^) that added piece is ⊥n^, so its dot with n^ is 0 and the shadow length does not change.
Why does the topic need this and not something else? The whole point of "shortest stick" is that it points perpendicular to both wires — and the cross product is the only standard tool that manufactures a both-perpendicular direction directly. See Vector Cross Product for the full determinant recipe.
That is the entire reason the main formula has ∣b1×b2∣ in the denominator: it is the "divide by its length" step that converts n=b1×b2 into the unit direction n^.
Why the topic loves it: the numerator of the main formula, (a2−a1)⋅(b1×b2), is a triple product [a2−a1,b1,b2]. And a box has zero volume exactly when its three edges lie flat in one plane — so:
The diagram below shows how each foundation feeds into the final shortest-distance formula. Read arrows as "is needed for". A point plus a direction build a line. The length bars and the cross product together build the unit vectorn^. The dot product turns n^ into a projection, and dot-plus-cross builds the scalar triple product. Those two strands — projection and triple product — join at the bottom into the shortest-distance result.
