I assume that you are familiar with vector and cross product in three dimension.

A line is defined by two points x_1, x_2:

x==x_1+(x_2-x_1)*s

where s is a variable.

The normal vector of skew lines l_1 (x_1,x_2) and l_2 (x_3,x_4) is the cross product of vectors (x_2-x_1) and (x_4-x_3)

N==(x_2-x_1)x(x_4-x_3)

and normalize it, we get

n==(x_2-x_1)x(x_4-x_3)/|(x_2-x_1)x(x_4-x_3)|

so the distance between line l_1 and line l_2 is

d==n.(x_3-x_1)

Easy? For more informations about vector and geometry, please refer to

http://mathworld.wolfram.com/topics/Geometry.html

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