<s>[INST] Why do we divide by $ \sqrt{A^2 + B^2} $ to convert a function from the standard form to the normal form? To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ \cos \omega \ x + \sin \omega \ y = p $, we first divide by $ \sqrt{A^2 + B^2} $. Why? 
For example, to convert the equation $ \sqrt 3 \ x + y - 8 = 0 $, we divide by $ \sqrt{A^2 + B^2} = 2 $ to get $ \frac{\sqrt 3}{2} \ x + \frac 12 \ y = 4 $ and then solve for $ \cos \omega = \frac{\sqrt 3}{2} $ and $ \sin \omega = \frac 12 $to find the value of $ \omega $. 
I understand that we can't possibly convert this directly, since there is no $ \omega $ for which $ \cos \omega = \sqrt 3 $--but why do we specifically divide by $ \sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ \cos \omega $ and $ \sin \omega $ between 0 and 1 and then solve it?
 [/INST] Given any two numbers $A$ and $B$ (not both equal zero), the numbers $a=A/\sqrt{A^2+B^2}$ and $b=B/\sqrt{A^2+B^2}$ satisfy $a^2+b^2=1$, so $a$ and $b$ are, respectively, the cosine and the sine of some angle. You are just finding a unit normal vector out of the given normal vector $(A,B)$ in the equation of the line. The same can be done in any number of dimensions: you can transform a given non-zero normal vector into a unit normal vector.
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