How Do You Calculate the Shortest Distance Between Two Skew Lines in Space?

[ThankYou]

1. The problem statement,
all variables and given/known data
There are 4 vectors a,b,u,v in space and {u,v} are linearly independen
The groups A={a+s*u|s E R} B = {b+t * v|t E R}are two lines in space

The Two A , B lines does not meet and the "part/line/segment (I don't know how it called ) " PQ is the smallest part that is ends are Q \in B ,P \in A show that the length of PQ is"
\frac{|(u \times v)\bullet(a-b)|}{||u \times v||}

Homework Equations

all calc/..

The Attempt at a Solution

I mange to get to the point of
u \times v =(s)
and then
\frac{L(s)*L(a-b)*cos(w)}{L(s)}=L(a-b)*cos(w)
bUT I don't know what's next

 

[gabbagabbahey]

I mange to get to the point of
u \times v =(s)
and then
\frac{L(s)*L(a-b)*cos(w)}{L(s)}=L(a-b)*cos(w)
bUT I don't know what's next

Huh? How are you managing to get to this point? What do w and L represent?

What is the general formula for the distance between two points \textbf{r}_1 and \textbf{r}_2? What if the points lie on the lines given in the problem (i.e. \textbf{r}_1=\textbf{a}+s\textbf{u} and \textbf{r}_2=\textbf{b}+t\textbf{v})? How would you go about minimizing that distance?

 

[ThankYou]

Yes I've managed to do it...
I've done:
u \times v and got the vector that is vertical to both u , v
Then I've made a plane that use this vector and use point a, then I've used the forumala of plane distance from point in the first line.
Thank you/