Distance between two lines in 3D

[Melawrghk]

Homework Statement

L₁=(x,y,z)=(-2,-4,-3)+t(-2,3,-2)
L₂=(x,y,z)=(0,0,4)+ s(-2,-3,-1)

Find the shortest distance between the two lines

The Attempt at a Solution

First, I did the cross product of the two directional vectors from both lines:
(-2,3,-2) x (-2,-3,-1) = (-9,-2, 12)
And its unit vector is:
-9x-2y+12z
-----------
229^1/2

Then I found the distance between two points on the lines:
R = (0,0,4)-(-2,-4,-3) = (2,4,7)

And then I just tried to project the R onto the unit vector and got 58/sqrt(229), but that's not right. Help?

 

[rock.freak667]

that should be correct.

EDIT: yeah Dick is right...Sorry about that

 

[Dick]

I get that the cross product is (-9,+2,+12).

 

[Melawrghk]

Ooohhh! I see, I'm just retarded and forgot about the negative space in the second row. OOOOH. This all makes sense now!

Thank you!