Question

How do you find the distance betweeen two parallel lines in `3` dimensional space?

Answer 1

One way...

Explanation:

If you have two parallel lines in `3` dimensional space then they can be described parametrically by:

`l_1(t) = bb(p) + t bb(v)`

`l_2(t) = bb(q) + t bb(v)`

where `bb(p) = (p_1, p_2, p_3)` and `bb(q) = (q_1, q_2, q_3)` are points through which the two lines pass, `bb(v) = < v_1, v_2, v_3 >` is the vector describing their common direction and `t` is a real parameter.

The plane:

`v_1 x + v_2 y + v_3 z = 0`

is normal to both lines and each line intersects it in exactly one point.

For the line `l_1`, the point of intersection satisfies:

`(bb(p)+t bb(v)) * bb(v) = 0`

Hence:

`t = -(bb(p) * bb(v)) / (bb(v) * bb(v))`

and the point of intersection is:

`bb(p) - ((bb(p) * bb(v)) / (bb(v) * bb(v))) bb(v)`

Similarly, the point of intersection of `l_2` with the normal plane is:

`bb(q) - ((bb(q) * bb(v)) / (bb(v) * bb(v))) bb(v)`

The distance between the lines is the distance between these two points:

`sqrt((bb(q)-bb(p)-(((bb(q)-bb(p)) * bb(v)) / (bb(v) * bb(v))) bb(v)) * (bb(q)-bb(p)-(((bb(q)-bb(p)) * bb(v)) / (bb(v) * bb(v))) bb(v)))`

Answer 2

Distance: `(|bbu xx (bbq-bbp)|)/absbbu`

Explanation:

Note: This formula only works for finding the shortest distance between two parallel lines

Finding the (shortest) distance between two parallel lines is the same as finding the distance between a line and point.

Let the line `l` going through the point `P` with position vector `bbp` in the direction of `bbu` have equation `bbr=bbp + lambdabbu`. Let `Q` with position vector `bbq` be a point on the line parallel to `l` and `M` be the point on `l` closest to `Q`.

Consider the triangle `PQM` having an interior angle of `vartheta` between `PM` (ie., `bbu`) and `PQ` (ie., `bbq-bbp`).

Then

`abs(bb(MQ))=abs(bbp-bbq)sinvartheta=(|bbu xx (bbq-bbp)|)/absbbu`

Remark: If `vartheta>pi"/"2`, then the expression remains the same as `sin(pi-vartheta)=sinvartheta`

Answer 3

See below

Explanation:

Given two parallel lines `L_1,L_2`

`L_1-> p = p_1 + lambda_1 vec v`
`L_2->p=p_2+lambda_2 vec v`

with `abs(vec v) = 1`

then

`d^2 = norm(p_2-p_1)^2- << p_2-p_1, vec v >> ^2`

then

`d = sqrt(norm(p_2-p_1)^2- << p_2-p_1, vec v >> ^2)`

NOTE

`<< cdot, cdot >>` represents the scalar product of two vectots