(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Closest possible points on skew lines

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