3D Geometry

The general method is perhaps best illustrated by working through an
example.

Find the shortest distance between the given lines, and the points of
closest approach on each line.

x y-3 z x-5 y-8 z-2
--- = --- = --- = s and ----- = ----- = ------ = t
1 1 -1 3 7 -1

The common perpendicular is obtained from the vector product

| i j k | = i(6) -j(2) + k(4)
| 1 1 -1 |
| 3 7 -1 |

So common perpendicular is the vector (3, -1, 2) which can be written
as a unit vector in the form 1/(sqrt(14)[3, -1, 2]

The vector connecting the given point on line (1) with the given point
on line (2) is [(5-0), (8-3), (2-0)] = (5, 5, 2)

The scalar product of this with the common perpendicular in unit
vector form is

5 x 3 + 5 x (-1) + 2 x 2 14
------------------------- = ------- = sqrt(14)
sqrt(14) sqrt(14)

So the shortest distance is sqrt(14).

Now to find the points of closest approach, we have:

line (1) is r = (0, 3, 0) + s(1, 1, -1) = [s, (3+s), -s]
line (2) is r = (5, 8, 2) + t(3, 7, -1) = [(5+3t), (8+7t), (2-t)]

and we must find s and t.

Line joining a general point on line(1) to a general point on line(2)
is the vector

[(5+3t-s), (8+7t-3-s), (2-t+s)]

[(5+3t-s), (5+7t-s), (2-t+s)]

and if s and t are points of closest approach this must be parallel to
the vector (3, -1, 2)

5+3t-s 5+7t-s 2-t+s
so ------ = -------- = ------
3 -1 2

From these equations s = -1 and t = -1

The points of closest approach are therefore

on line(1) (-1, 2, 1) and on line(2) (2, 1, 3)

Check that shortest distance is sqrt(14)

shortest distance = sqrt((2+1)^2 + (1-2)^2 + (3-1)^2)

= sqrt(9 + 1 + 4) = sqrt(14)

If you require the mid-point of these points, you get

-1+2 2+1 1+3
----, ----, ---- = [1/2, 3/2, 2]
2 2 2