Closest possible points on skew lines

[lubricarret]

Homework Statement

Find points P,Q which are closest possible with P lying on line:
x=7-5t, y=-5+11t, z=-3-1t
and Q lying on line:
x=-354-8t, y=-194+12t, z=-73+7t

*the line joining P + Q is perpendicular to the two given lines.

Homework Equations

Projection formula, cross product...

The Attempt at a Solution

So, this is the first time I've seen a problem with skew lines, so am a bit confused how to go about this one.

I wrote the equations to be :
X_p = [7,-5,-3] + [-5,11,-1]s
X_q = [-354,-194,-73] + [-8,12,7]t

Therefore, the direction vectors are:
d_p = [-5,11,-1] and
d_q = [-8,12,7]

I took the cross product: [-5,11,-1] x [-8,12,7] to get the normal that is perpendicular to both lines, to be:
[89,43,28]

I figured I would take a point on Line P, and a point on Line Q, to get an arbitrary vector connecting the two lines; I took the points given in the equation:
[7,-5,-3] - [-354,-194,-73] to get v = [361,-189,70]

I then projected this onto the normal, so took:
proj_n_v


Am I correct in doing this? Would this proj_n_v be the distance from P to Q? If it is, what do I do now to get the two points P and Q.

Am I going about this correctly; I've never encountered this type of problem before, but see it in the practice problems on projections...


Thanks so much.

 

[Dick]

Yes, the projection will give you the distance. To get the actual two points, write W=[89,43,28] (your cross product). Then the difference of the two points must be parallel to W. So write X_p(s)-X_q(t)=u*W. That gives you three equations in three unknowns, s,t and u. Solve them. You could also do this more directly by minimizing (X_p(s)-X_q(t))^2. (Take the two partial derivatives wrt s and t and set them to zero.).

 

[lubricarret]

Thanks for the reply Dick!

Ahh k, got it now! )