1. The problem statement, all variables and given/known data 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. 2. Relevant equations Projection formula, cross product.... 3. 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.