Need some clarification on skew vectors

So we have two vectors in 3d space - they aren't parallel.

Line A goes through (1,2,7) and (2,3,4). Line B goes through (2,3,7) and (4,1,2).

I worked out the equations to be

r1=(1,2,7)+t(1,1,-3)
r2=(2,3,7)+s(2,-2,-5)

The question asks for me to find the shortest distance between these two lines... and it gives a hint: Find any vector that joins a point from one line to the other and then compute the scalar projection of this vector onto the vector orthogonal to both lines.

So I found a vector which I called PQ that goes through (1,2,7) and (2,3,7) which i get as (1,1,0).

For the vector that is orthogonal to both lines... I know I can use the cross product to find this but since I have parameters t and s, I'm not sure how to do it... Do I have to find the cross product between the direction of vectors r1 and r2?

Finally, I simply don't understand how the scalar projection of the vector PQ onto the vector perpendicular to both lines will give me the length. I understand the scalar projections in 2d vectors when they are tail to tail but the vectors PQ and the perpendicular vector aren't even touching, so how does that work?

Thanks in advanced!

Quote:

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Originally Posted by jgv115

Line A goes through (1,2,7) and (2,3,4). Line B goes through (2,3,7) and (4,1,2).
I worked out the equations to be
r1=(1,2,7)+t(1,1,-3)
r2=(2,3,7)+s(2,-2,-5)

The question asks for me to find the shortest distance between these two lines

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Given two skew lines the distance between them is

Jgv115

I have tried the vectorial equations of the lines r1 and r2 and I have verified that they are skew lines ( they do not intercept in 3D).
you have therefore 2 skew lines in space and according to a well known theorem of 3D Eucledian Geometry there is a unique pair of parallel planes that contain these two lines..
Lets find the equation of one of them that contains the line r1. you need the normal vector and the only way to find this is to get the vector product of the two vectors that are parallel to the lines r1 and r2. this product yields n(-11,-1,-4) ( verify pls).
therefore the plane that includes the line r1 and is parallel to the line r2 has equation 11x+y+4z=0
now get the distance of any point of the line r2 to this plane and this will solve your problem

the distance of the point 2,3,7 which lies on the line r2 from the plane we just found is d=(11 x 2 +1 x 3+7 x 4)/(sqroot(11^2+1^2+4^2))
and thus d=4.511
Please verify the calculations

MINOAS