Perpendicular distance between two skew lines?

[math12345]

Homework Statement

The lines r=(2+2t)i+(3-3t)j+(4-2t)k and
v=(4+2s)i+(1-6s)j+(5-3s)k are skew. Find the perpendicular distance between them at the point(s) where they cross over each other.

Homework Equations

The Attempt at a Solution

Does this seem right?
parametrics for r:
x=2+2t
y=3-3t
z=4-2t
parametrics for v:
x=4+2s
y=1-6s
z=5-3s

u=<2,-3,-2>
v=<2,-6,-3>

cross product of u and v is n= <-3,2,-6>
points p(2,3,4) and q(4,1,5)

<-3,2,-6>*<x-2,y-3,z-4>=0
-3(x-2)+2(y-3)-6(z-4)=0
-3x+6+2y-6-6z+24=0
-3x+2y-6z+24=0

d=abs(-3*4+2*-6*5)/sqrt(-3)^2+(2)^2+(-6)^2
=abs(-12+2-30)/sqrt49

final answer: 40/7?

 

[LCKurtz]

Homework Statement

The lines r=(2+2t)i+(3-3t)j+(4-2t)k and
v=(4+2s)i+(1-6s)j+(5-3s)k are skew. Find the perpendicular distance between them at the point(s) where they cross over each other.

Homework Equations

The Attempt at a Solution

Does this seem right?
parametrics for r:
x=2+2t
y=3-3t
z=4-2t
parametrics for v:
x=4+2s
y=1-6s
z=5-3s

u=<2,-3,-2>
v=<2,-6,-3>

cross product of u and v is n= <-3,2,-6>
points p(2,3,4) and q(4,1,5)

All OK to here. I don't follow the rest and it isn't correct. All you need to do now is
let V = q-p = <2,-2,1>. Then to get the distance between the lines, take the absolute value of the component of V along N:

d = | V dot N/length(N)|

<-3,2,-6>*<x-2,y-3,z-4>=0
-3(x-2)+2(y-3)-6(z-4)=0
-3x+6+2y-6-6z+24=0
-3x+2y-6z+24=0

d=abs(-3*4+2*-6*5)/sqrt(-3)^2+(2)^2+(-6)^2
=abs(-12+2-30)/sqrt49

final answer: 40/7?