Skew lines — shortest distance: Question bank

Quick vocabulary refresh so every symbol below is earned:

$L_{1} : r = a_{1} + λ b_{1}$ means "start at the fixed point $a_{1}$ , then walk any distance $λ$ (a plain number, positive or negative) in the direction $b_{1}$ ." See Equation of a Line in 3D.
$L_{2} : r = a_{2} + μ b_{2}$ is the second line, read the same way: start at $a_{2}$ , walk distance $μ$ in direction $b_{2}$ . Here $λ$ and $μ$ are just the two "sliders" that pick a point on each line.
$b_{1} \times b_{2}$ is the cross product: a new vector that stands at a right angle to both $b_{1}$ and $b_{2}$ at once. We name this direction $n = b_{1} \times b_{2}$ (the raw normal), and its unit-length version $\overset{n}{^} = n /∣ n ∣$ (the unit normal, length exactly $1$ ).
$[u, v, w] = u \cdot (v \times w)$ is the scalar triple product: one number whose size is the volume of the box built from the three vectors.

True or false — justify

Two lines that never meet must be skew.

False — parallel lines also never meet, yet they are not skew because skew additionally requires different directions. "Never meet" splits into two cases: parallel and skew.

b_{1} \times b_{2} = 0

then the shortest-distance formula still applies, you just get a small number.

False — a zero cross product means the denominator is

0

, so the fraction is undefined (

0/0

or divide-by-zero), not "small." You must switch to the parallel-line formula.

The shortest segment between two skew lines is perpendicular to both of them.

True — at the minimum, sliding either endpoint along its line gives no first-order shortening, and "no change while moving along

b

" means the segment is perpendicular to

b

Changing which point

a_{1}

I pick on

L_{1}

changes the computed distance.

False — moving

a_{1}

adds a multiple of

b_{1}

to the gap, and

b_{1} \cdot (b_{1} \times b_{2}) = 0

, so the projection onto the common perpendicular is untouched.

If the scalar triple product

[a_{2} - a_{1}, b_{1}, b_{2}] = 0

, the lines are definitely skew.

False — a zero triple product means the lines are coplanar, so (given they are not parallel) they intersect. Coplanar non-parallel is the opposite of skew.

The distance formula can output a negative number, so we take the absolute value.

True in spirit — the triple product (numerator before

∣ \cdot ∣

) can be negative because it is a signed volume; distance

d

must be

\geq 0

, so the absolute value is essential, not decorative.

Two lines in the same plane are always intersecting.

False — coplanar lines are either intersecting or parallel. Being coplanar rules out skew, but does not force a meeting point.

If two lines have the same direction vector they cannot be skew.

True — equal (or proportional) directions make them parallel, and parallel is a separate category from skew by definition.

The cross product

b_{1} \times b_{2}

and

b_{2} \times b_{1}

give different shortest distances.

False — they differ only by a sign, and since the final formula takes an absolute value of the numerator and a magnitude in the denominator, the distance is identical.

Spot the error

"Distance

= \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} ∣∣ b _{2} ∣}

The denominator is wrong — it must be

∣ b_{1} \times b_{2} ∣

, the magnitude of the cross product, not the product of the two magnitudes. These agree only when

b_{1} ⊥ b_{2}

, so the mistake hides in perpendicular cases.

"The lines don't share an obvious common point, so they are skew."

Missing a check — you must compute the triple product. If it is

0

(and directions differ), they secretly intersect; "no obvious point" is not proof.

"For parallel lines I'll just use

b_{1} \times b_{2}

in the numerator."

For parallel lines

b_{1} \times b_{2} = 0

, killing both numerator and denominator. Use the parallel formula

d = \frac{∣ b \times ( a _{2} - a _{1} ) ∣}{∣ b ∣}

instead — the point-to-line distance.

"I got a negative triple product, so I made an arithmetic mistake."

Not necessarily — the raw triple product is a signed volume, so negative is perfectly normal. Only the final distance

d

must be non-negative, handled by the absolute value.

"The shortest stick connecting the lines is perpendicular to

L_{1}

, so any such perpendicular works."

It must be perpendicular to both lines simultaneously. There are infinitely many perpendiculars to a single line; only the common perpendicular (direction

n = b_{1} \times b_{2}

) gives the true shortest distance.

"Since

d = 0

, the lines coincide (are the same line)."

d = 0

with different directions means they cross at one point, not that they coincide. Coinciding requires identical direction and a shared point.

"I'll compute

d = ∣ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) ∣

directly."

Forgot to divide by

∣ b_{1} \times b_{2} ∣

. The dot product with the un-normalised

n

overcounts by the length of

n

; you must project onto the unit vector

\overset{n}{^}

, i.e. divide by

∣ n ∣

Why questions

Why does the cross product, not the dot product, give the direction of the shortest segment?

The cross product is defined to be perpendicular to both inputs, which is exactly the "right-angle-to-both-lines" direction

n

we need; the dot product only outputs a number, not a direction.

Why does the shortest distance

d

not depend on the sliders

λ

μ

Those parameters just slide points along the lines, adding multiples of

b_{1}, b_{2}

to the gap; both are perpendicular to

\overset{n}{^}

, so they vanish under the projection.

Why must skew lines live in different planes?

If a single plane contained both, they would be coplanar and hence either meet or be parallel — neither of which is skew. Skew is precisely "no common plane exists."

Why is the numerator a scalar triple product?

(a_{2} - a_{1}) \cdot (b_{1} \times b_{2})

has exactly the form

u \cdot (v \times w)

, whose absolute value is the volume of the parallelepiped built from the gap and the two directions — see Scalar Triple Product.

Why does a zero volume (triple product

= 0

) mean the lines meet?

Zero volume means the three vectors are flat in one plane, so both lines share that plane; coplanar non-parallel lines must cross. This is the Coplanarity of Two Lines test.

Why can the same formula be written in Cartesian determinant form?

The determinant with rows

(a_{2} - a_{1}), b_{1}, b_{2}

is the triple product expanded into components, and the square-root denominator is

∣ b_{1} \times b_{2} ∣

in components — identical maths, different notation.

Why do we divide by

∣ b_{1} \times b_{2} ∣

specifically?

To convert the projection onto the raw normal

n = b_{1} \times b_{2}

into a projection onto the unit normal

\overset{n}{^}

; dividing by

∣ n ∣

turns "how far along

n

" into a true length.

Edge cases

The two direction vectors are perpendicular (

b_{1} \cdot b_{2} = 0

). Does the formula still work?

Yes — perpendicular directions are still non-parallel, so

b_{1} \times b_{2} \neq = 0

and the standard formula applies unchanged; nothing special about

9 0^{\circ}

One line passes through the origin, so

a_{1} = 0

. Any trouble?

None — the gap simply becomes

a_{2} - 0 = a_{2}

. The origin is just a convenient point on

L_{1}

; the formula never cared which point we chose.

The lines are parallel and the point of one lies on the other. What is

d

d = 0

: the lines coincide (they are the same line). The point-to-line distance from a point already on the line is zero.

The gap vector

a_{2} - a_{1}

is itself parallel to

n = b_{1} \times b_{2}

. What does that mean geometrically?

The chosen connecting points already lie along the common-perpendicular direction, so the whole gap projects onto

\overset{n}{^}

— the distance

d

equals

∣ a_{2} - a_{1} ∣

and those points are the feet of the shortest segment.

Both lines pass through the same single point but point differently. Skew or not?

Intersecting, not skew — a shared point forces the triple product to

0

(the gap can be taken as

0

), so

d = 0

The direction vectors are anti-parallel, e.g.

b_{2} = - 3 b_{1}

. Standard formula or special case?

Special case — anti-parallel still means proportional, so

b_{1} \times b_{2} = 0

; treat them as parallel and use the point-to-line formula.

Recall One-line self-test

Given non-parallel lines with triple product exactly $0$ , name their relationship. ::: They intersect — coplanar plus non-parallel means they cross at one point.