I didn't explain very well what I did.
lines
1-2: define Vn as a quaternion vector, and x being a real coefficient.
2-4: define functions f(x) and g(x) that they equal your two lines.
4-6: find Vi such that V2xVi=i, which can be found by premulitiplying by V2^-1
6-10: dilate both f(x) and g(x) by a factor of Vi, then translate both functions. This means that one function is the I axis (xi) and the other is now just a jumble. This is where my stuff falls apart, by multiplying g(x) by Vi, you introduce a real part into the system and g(x)xVi, which makes everything turn to shit.
10-13: What i try to do here is shift the g(x) variant along the real axis (which shouldn't change the line at all, because it's imaginary, but it does because I fucked up.) and remove the i term from the constant. Vs is basically the i intercept of the line.
The distance now, should be the j,k distance away from the i axis. so basically, the distance from the origin to the point (x,y).
This next step I forgot to add:
you can now differentiate the difference, find the turning point, and then substitute the x into the
original line You have dilated the lines, so you need to use the original lines to find the distance. use the i coefficient of h(x) to substitute into f(x). Then find the distance between f(x) and g(x) at the two points found.
question easily done
EDIT: I get your reasoning, both planes will be parallel, even if both vectors are parallel (however, then there will only be two lines not planes). therefore if you find a line perpendicular to the planes, and find the intersections between the lines and the planes, the distance between the two will be the shortest distance between the two lines.
EDIT2: here is a general (working) solution. by substituting 1/c(x+ck) you get an equation of the form below, with no k therm in the constant, and 1 as the coefficient of k in the other term. This does not change the line shape at all.