Linear Algebra: Distance between two parallel lines Find the distance between the two (obviously parallel) lines below where $\alpha ,\beta \in \mathbb R$ are scalars.
$$\text{Line #1}:\begin{bmatrix}0\\1\\2\end{bmatrix}+\alpha\begin{bmatrix}1\\1\\0\end{bmatrix},\text{Line #2}:\begin{bmatrix}1\,\\-1\\3\,\end{bmatrix}+\beta\begin{bmatrix}1\\1\\0\end{bmatrix}$$
How do I find the required distance? Please give me advice.
 A: Pick point on the first line.
$P_{1}= [1,2,2]$
Pick point with parameter $\beta$.
$P_{2}= [1,-1,3] + \beta[1,1,0]$
Make a function   $ f$($\beta$) which describes distance between these two points.
$ f(\beta) = $ $\sqrt[]{ (1-1-\beta)^2 + (2+1-\beta)^2 + (2-3)^2 }$
Find minimum of this function.
min$f(\beta) = \sqrt{11\over2}$
So the distance is $\sqrt{11\over2}$
$$$$$$$$
--- how to find minimum: 
square function is strictcly monotonic, and values of function under square root $\in R+$  so we can find min of 
$ f(\beta) = { (1-1-\beta)^2 + (2+1-\beta)^2 + (2-3)^2 }$
simplify
$ f(\beta) =  2\beta^2 - 6\beta + 10 $
this function has minimum which is y of the vertex of parabola
y = $[-\Delta]\over{4a}$
$\Delta = b^2 - 4ac$
so we have $\Delta= 36-4\times 2 \times 10 = -44$
min f($\beta$)=$\sqrt{y}$= $ \sqrt{44\over 8} $=$ \sqrt{11\over2}$
A: If you know the angle between your lines and $OX$, you can rotate the lines to make them parralel with $OX$ or $OY$. 
After this, just calculate the difference between the cordinates.
