Visual walkthrough — Skew lines — shortest distance

---

Step 1 — Two wires that never meet

WHAT. Picture two perfectly straight wires floating in a room. Wire 1 passes through a point we call ${\vec{a}}_1$ and runs in the direction of an arrow ${\vec{b}}_1$. Wire 2 passes through ${\vec{a}}_2$ and runs along ${\vec{b}}_2$.

Every point on wire 1 is ${\vec{a}}_1+\lambda {\vec{b}}_1$: start at ${\vec{a}}_1$, then walk $\lambda$ steps along ${\vec{b}}_1$. Here $\lambda$ is just a dial — turn it and you slide along the wire. Same story for wire 2 with its own dial $\mu$.

WHY. We need names for "where a wire is" ($\vec{a}$) and "which way it runs" ($\vec{b}$) before we can measure anything. The dials $\lambda ,\mu$ let us talk about any point on each wire at once.

PICTURE. The blue wire and the orange wire pass at different heights — like two overpasses. They never touch, and they don't point the same way. That is exactly what skew means.

---

Step 2 — The joining stick, and why we want it short

WHAT. Pick any point $P$ on wire 1 and any point $Q$ on wire 2. Draw the arrow $\overrightarrow{PQ}$ from $P$ to $Q$. Its length $|\overrightarrow{PQ}|$ is a distance between the wires — but not usually the shortest one.

WHY. "Shortest distance" is a minimum. To find a minimum we first need a quantity to minimise, and that quantity is the length of this joining stick as we slide $P$ and $Q$ around.

PICTURE. Watch three different joining sticks. The gray ones are long and lie at a slant. There is one special green stick that is shortest — our whole job is to pin it down.

---

Step 3 — Why the shortest stick pokes both wires at a right angle

WHAT. Claim: at the shortest stick, $\overrightarrow{PQ}$ is perpendicular to wire 1 and to wire 2.

WHY. Suppose the stick met wire 1 at a slant (not a right angle). Then $\overrightarrow{PQ}$ is the hypotenuse of a little right triangle: one leg lies along the wire, the other drops straight across. A hypotenuse is always longer than a leg — so by sliding $P$ a hair along the wire we could shorten the stick. If we can still shorten it, it wasn't the shortest. So at the true minimum, no slide helps, which forces the right angle on both wires.

PICTURE. The red slanted stick is a hypotenuse; the green perpendicular stick is its shorter leg. Sliding $P$ toward the foot of the perpendicular shrinks the stick until it stands square.

---

Step 4 — One direction is perpendicular to both: the cross product

WHAT. We need a single arrow that is at right angles to ${\vec{b}}_1$ and ${\vec{b}}_2$ simultaneously. That arrow is the cross product $\vec{n}=\underset{\text{dir of wire 1}}{\underbrace{{\vec{b}}_1}}\times \underset{\text{dir of wire 2}}{\underbrace{{\vec{b}}_2}}.$

WHY this tool and not another? We asked a very specific question: "give me a direction square to two given arrows." The cross product is defined to answer exactly that — its output is, by construction, perpendicular to both inputs. No other single operation on two vectors hands you a perpendicular direction for free. (A dot product answers a different question — "how aligned are two arrows?" — and returns a number, not a direction, so it can't do this job.)

Term by term: $\vec{n}$ is the new perpendicular arrow; the $\times$ says "build the vector square to both"; its length $|\vec{n}|$ measures the area of the parallelogram the two $\vec{b}$'s span (bigger when they point more differently).

PICTURE. The green $\vec{n}$ shoots straight out of the flat sheet spanned by ${\vec{b}}_1$ and ${\vec{b}}_2$, like a flagpole out of a tabletop.

---

Step 5 — Shrink $\vec{n}$ to a unit arrow $\hat{n}$

WHAT. We only want $\vec{n}$'s direction, not its length. So divide it by its own length to get a pointer of length exactly $1$: $\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{{\vec{b}}_1\times {\vec{b}}_2}{|{\vec{b}}_1\times {\vec{b}}_2|}.$

WHY. In the next step we measure "how far the gap reaches in the $\vec{n}$ direction." A measuring pointer must have length $1$, otherwise its own length would contaminate the reading. The little hat $\hat{}$ always means "same direction, length trimmed to $1$."

Term by term: top $\vec{n}$ = the raw perpendicular arrow; bottom $|\vec{n}|$ = its length; the quotient = pure direction, size $1$.

PICTURE. The long green $\vec{n}$ becomes a short unit arrow $\hat{n}$ pointing the same way — a clean ruler aimed along the common perpendicular.

---

Step 6 — The gap arrow, and reading its shadow with the dot product

WHAT. Connect the two anchors: the gap arrow is ${\vec{a}}_2-{\vec{a}}_1$ (from wire 1's anchor to wire 2's anchor). The shortest distance is how much of this gap points along $\hat{n}$: $d=|({\vec{a}}_2-{\vec{a}}_1)\cdot \hat{n}|.$

WHY the dot product? New question: "how much of arrow A points along direction B?" The dot product answers exactly this — $({\vec{a}}_2-{\vec{a}}_1)\cdot \hat{n}$ returns the length of the gap's shadow cast onto the $\hat{n}$ line. That shadow is the shortest distance, because moving along either wire only adds $\vec{b}$-arrows, which are perpendicular to $\hat{n}$ and cast zero shadow.

Term by term:

• ${\vec{a}}_2-{\vec{a}}_1$ = the gap arrow between anchors;
• $\cdot \hat{n}$ = "take the shadow along the perpendicular";
• $|\cdot |$ = a distance can't be negative, so we keep the size only.

PICTURE. The gap arrow slants across; its shadow onto the $\hat{n}$ ruler is the green segment — and that green length is $d$.

---

Step 7 — Assemble the finished formula

WHAT. Substitute $\hat{n}=\frac{{\vec{b}}_1\times {\vec{b}}_2}{|{\vec{b}}_1\times {\vec{b}}_2|}$ into $d=|({\vec{a}}_2-{\vec{a}}_1)\cdot \hat{n}|$: $\boxed{d=\frac{|({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)|}{|{\vec{b}}_1\times {\vec{b}}_2|}}$

WHY. The unit-hat's denominator floats up to become the formula's denominator. The top is a scalar triple product $[{\vec{a}}_2-{\vec{a}}_1,{\vec{b}}_1,{\vec{b}}_2]$ — geometrically the volume of the box built on those three arrows; the bottom is the area of its base. Volume ÷ base area = height, and the height of that box is exactly the perpendicular gap between the wires.

Term by term: numerator = box volume (gap along perpendicular, un-normalised); denominator = base parallelogram area; ratio = height = $d$.

PICTURE. The three arrows frame a slanted box (a parallelepiped). Volume over base equals height — the green vertical stick.

---

Step 8 — The degenerate case: parallel wires

WHAT. If ${\vec{b}}_1\parallel {\vec{b}}_2$ (call the shared direction $\vec{b}$), then ${\vec{b}}_1\times {\vec{b}}_2=\vec{0}$ and the boxed formula becomes $\frac{0}{0}$ — undefined. Switch to: $d=\frac{|\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)|}{|\vec{b}|}.$

WHY. Parallel wires are like two rails: the shortest distance is just the perpendicular drop from any point of one rail to the other — a point-to-line distance. The cross product $\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)$ has length = (area of the parallelogram on $\vec{b}$ and the gap); dividing by $|\vec{b}|$ (the base) again gives the height — the perpendicular rail-to-rail gap.

PICTURE. Two parallel rails; the perpendicular drop is the same everywhere, so pick any convenient point.

---

The one-picture summary

Every step compressed into a single frame: two wires, the cross product raising a flagpole $\vec{n}$ square to both, the gap arrow casting its shadow along $\hat{n}$, and the slanted box whose volume ÷ base = height is the answer.

Recall Feynman retelling of the whole walkthrough

Two straight wires float in a room. They never touch and don't point the same way. I want the length of the shortest stick joining them. First I notice the shortest stick must hit both wires dead square — because if it hit at a slant, it'd be a hypotenuse and I could always find a shorter leg by sliding along the wire. So I need the one direction that's square to both wires: the cross product ${\vec{b}}_1\times {\vec{b}}_2$ hands me that for free (that's literally what it's built to do), and I shrink it to a length-1 pointer $\hat{n}$. Now I connect any point of wire 1 to any point of wire 2 — that's my gap arrow — and I ask "how much of it points along $\hat{n}$?" The dot product reads off that shadow, and the shadow is the distance, because sliding along either wire only adds sideways pieces that cast no shadow. Written as one boxed formula it's just the volume of the slanted box on the three arrows, divided by the box's base area — volume over base is height, and the height is my stick. If the wires happen to be parallel, the cross product collapses to zero, so instead I just drop a perpendicular from one rail to the other.

Recall Quick self-check

Why perpendicular to both wires? ::: A slanted stick is a hypotenuse; you can always slide to a shorter leg, so the true minimum has no slack — a right angle on both. Why the cross product and not the dot product to get $\vec{n}$? ::: The cross product outputs a direction square to two arrows; the dot product outputs a number measuring alignment — wrong kind of answer. What does the numerator mean geometrically? ::: The volume of the parallelepiped (box) built on ${\vec{a}}_2-{\vec{a}}_1,{\vec{b}}_1,{\vec{b}}_2$. What does dividing by $|{\vec{b}}_1\times {\vec{b}}_2|$ do? ::: Divides box volume by base area, leaving the height = shortest distance. Why may I use any anchor points? ::: Sliding along a wire adds $\vec{b}$-pieces which are $\perp \hat{n}$ and cast no shadow, so the projection is unchanged. What if the triple product is $0$ (non-parallel lines)? ::: The box is flat, the four points are coplanar, so the lines intersect.

---

See also: Equation of a Line in 3D · Plane containing a line · parent topic.