# Thread: Shortest distance between 2 skew lines

1. ## Shortest distance between 2 skew lines

Given 2 skew lines

ru = (x1, y1, z1) + u(a1, b1, c1)
rv = (x2, y2, z2) + v(a2, b2, c2)

Verify the formula for the shortest distance of the line.

Okay what I did was that I found the distance between 2 points r = (x1-x2, y1-y2, z1-z2) and then generated a vector that is orthogonal to the 2 lines using cross product and projected r onto d (the distance). The problem I'm having is simplying the equation into the final equation. How do I simplify r.(u x v) into that 3x3 matrix?

2. ## Re: Shortest distance between 2 skew lines

Originally Posted by aqualiary
Given 2 skew lines
ru = (x1, y1, z1) + u(a1, b1, c1)
rv = (x2, y2, z2) + v(a2, b2, c2)
Verify the formula for the shortest distance of the line.
I can give you the usual formula.
Suppose that $\displaystyle P+tD~\&~Q+tE$ are two skew lines, then the distance between them is:
$\displaystyle \frac{|(P-Q)\cdot(D\times E)|}{\|D\times E\|}$.

Now the denominator of that expression is the length of $\displaystyle D\times E$.
Do you know how to find $\displaystyle D\times E$ using a determinate?

3. ## Re: Shortest distance between 2 skew lines

So basically I have (P-Q).(D X E) down to 3 2x2 matrices (each adding the other), how do you put that into a 3x3 form?

4. ## Re: Shortest distance between 2 skew lines

Originally Posted by aqualiary
So basically I have (P-Q).(D X E) down to 3 2x2 matrices (each adding the other), how do you put that into a 3x3 form?
The first row is the entries in $\displaystyle P-Q$.
The the next to rows are the usual $\displaystyle D\times E$.
That is known as triple product.