I'll start with a general description of how you can do this. Then I'll give a particular example.

We start with 2 skew lines which I'll call line L(1) and line L(2).

Suppose line L(1) has the vector equation of

This means that its direction vector is

Similarly, suppose line L(2) has the vector equation of

This means that its direction vector is

I think that the easiest way to picture, and so find the shortest distance
between these 2 lines, is through using the equation for a plane.

We will use the forms Ax + By + Cz = **r.n** = D where **n** is
a unit normal to the plane.

Then we have **n** = (A, B, C). Also, |D| gives the perpendicular
distance from the origin to the plane.

(You may need to go through my pages on planes to be happy with this. In
particular, you may need equations of planes in the
form Ax + By + Cz = D.)

Now, here's how we use planes to solve our problem.

Imagine sliding a copy of line L(2) up towards line L(1) until it
intersects with it at P.
It doesn't matter what path you take so long as you keep the direction of
L(2) unchanged.

Now, line L(1) and the shifted line L(2) will lie in a plane, P(1), and
the point P with position vector **p** lies on this plane.

Also, we can find a normal vector for P(1) by working out the cross product
of **d** and **e**. This can be adjusted to give a unit normal vector
for plane P(1) which I'll call **n**.

We can now write its vector equation
as **r.n** = **p.n** = **d**_{1} where
|**d**_{1}| is the perpendicular distance of plane P(1) from the
origin.

Similarly, we can shift a copy of line L(1) until it intersects with
line L(2) at Q. The
shifted line L(1) and line L(2) lie in a second plane P(2) with point Q
lying in it. The planes P(1) and P(2) are parallel to each other, and the
**n** already found is a unit normal vector to P(2) as well.

So the vector equation of plane P(2) can be written
as **r.n** = **q.n** = **d**_{2} where
|**d**_{2}| is the perpendicular distance of plane P(2) from the
origin.

The shortest distance between the 2 skew lines L(1) and L(2) is now
given by |**d**_{1} - **d**_{2}|.

The modulus sign matters here because it is taking care of whether the 2
planes are on the same side or opposite sides of the origin.

See my drawing below showing **p.n** = **d**_{1} and
**q.n** = **d**_{2} in the two cases.

In case (1), both

**A numerical example**

Find the shortest distance between the two skew lines L(1) and L(2) with
equations

**r** = (1, 2, 2) + s (4, 3, 2) and
**r** = (1, 0, -3) + t (4, -6, -1).

The point P with position vector **p** = (1, 2, 2) lies on L(1).

Shifting a copy of L(2) until it intersects with L(1) at P gives plane
P(1) with the point P lying in it.

The point Q with position vector **q** = (1, 0, -3) lies on L(2).

Shifting a copy of L(1) until it intersects with L(2) at Q gives plane
P(2) with the point Q lying in it.

The cross product of the two
direction vectors, (4, 3, 2) and (4, -6, -1) gives a common normal to both
planes.

Working this out gives the vector (9, 12, -36).

So a unit normal vector to both planes is given by 1/13 (3, 4, -12).

The equation of P(1) is
1/13 (3, 4, -12)** . r
** = 1/13 (3, 4, -12) **.** (1, 2, 2) = -1.

The equation of P(2) is
1/13 (3, 4, -12)** . r
** = 1/13 (3, 4, -12) **.** (1, 0, -3) = 3.

So the shortest distance between the 2 skew lines L(1) and L(2)
is |-1-3| = 4 units.

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