Distance between two 3D lines What is the distance between the 3D lines $x = \begin{pmatrix} 1 \\ 2 \\ -4 \end{pmatrix} + \begin{pmatrix} -1 \\ 3 \\ -1 \end{pmatrix} t$
and
$y = \begin{pmatrix} 0 \\ 3 \\ 5 \end{pmatrix} + \begin{pmatrix} 1 \\ 4 \\ 3 \end{pmatrix} u.$
Hint: To find the distance, try to calculate  $AB$'s smallest possible value, where $A$ represents a point which lies on one line and $B$ represents a point which lies on the other line.
I remember a handy formula here: http://math.harvard.edu/~ytzeng/worksheet/distance.pdf that allows mathematicians to find the distance between 3D lines, but I would rather solve the problem given the method that was hinted. 
I'm not entirely sure how to solve it this way, however. Would I try to find a value for both $t$ and $u$ that makes lines $x$ and $y$ the smallest value? I don't know if I'm thinking of solving this problem correctly.
Thanks.

Yes, to derive this, the natural approach would be to consider the function
$$ F(t,u) = |x(t)-y(u)| $$
If you plug the definition of $x(t)$ and $y(u)$ into the distance formula and simplify a bit, you get $F(t,u) = \sqrt{P(t,u)}$ where $P$ is some quadratic polynomial in two variables. The minimum of $F$ occurs at the same $(t,u)$ as the minimum of $P$, which should be easy to find using calculus. Then plug the $(t,u)$ you find back into the expression you got for $F$.
