Distance between two skew lines given by parametric equations (and more)

[Interesting]

Homework Statement

L1 L2
x= 1+t x= 1+2s
y= 1+6t y= 5+15s
z= 2t z= -2+6s

My professor said to find a normal vector and project the lines, but I'm new to this calculus and all these words are just a fuzzy cloud over my head.

Homework Equations

Projection of c onto L1 -> (C dot L1 / magnitude C) (L1 / magnitude L1)
Or (C dot L1)C/(magnitude C^2)

The Attempt at a Solution

L1xL2= a vector of <6,2,3> = C
The direction vector for L1 = <1,6,2>
L2 = <2,15,6>

So I think I need to project one onto the other...
But when I do that I get another vector, that's <144/7,48/7,72/7>. Do I find the magnitude of this, and that should be the distance? I'm lost. The magnitude of that vector is 24, which is a clean answer so it could be right...

 

[rootX]

Homework Statement

L1 L2
x= 1+t x= 1+2s
y= 1+6t y= 5+15s
z= 2t z= -2+6s

My professor said to find a normal vector and project the lines, but I'm new to this calculus and all these words are just a fuzzy cloud over my head.

Homework Equations

Projection of c onto L1 -> (C dot L1 / magnitude C) (L1 / magnitude L1)
Or (C dot L1)C/(magnitude C^2)

The Attempt at a Solution

L1xL2= a vector of <6,2,3> = C
The direction vector for L1 = <1,6,2>
L2 = <2,15,6>

So I think I need to project one onto the other...
But when I do that I get another vector, that's <144/7,48/7,72/7>. Do I find the magnitude of this, and that should be the distance? I'm lost. The magnitude of that vector is 24, which is a clean answer so it could be right...

You normal vector is wrong. If c is normal (perpendicular) to L1 or L2 then
c.L1 = 0
c.L2 = 0

 

[rootX]

But, I think here's what you are doing
http://img230.imageshack.us/img230/9503/64421739uy1.gif