The minimum number of variables to represent a 3D line in a unique way To my best understanding, the minimum number of variables to represent a line in 3D space is four. It means you need at least four values to identify a 3D line. For example from here 
$$a x + b y + c z = d \tag{1}$$
defines a line with four variables. However, this is not a unique representation of the line, and by scaling the equation you may still represent the same line:
$$k a x + k b y + k c z = k d \, , \{k \neq 0\} \tag{2}$$
So my question is that what is the minimum number of variables to represent a line in a unique way? In other words, if 
$$l_1 \equiv L\{ a_1, a_2, \cdots , a_n \} \tag{3}$$ 
and 
$$l_2 \equiv L\{ b_1, b_2, \cdots , b_n \} \tag{4}$$ 
then 
$$l_1 \equiv l_2 \tag{5}$$ 
only and if only 
$$ \{a_i = b_i, \forall i \in 1,\cdots,n\} \tag{6}$$
Or, if I want to ask my question differently, what is the best way to mathematically represent a line in a 3D space, in a unique way, with the minimum number of degrees of freedom? 
P.S.1. I think I have my answer, and it is shamefully simple. Just divide the first equation by $d$. So it seems the minimum number of variables to represent a line (or DOF) in a 3D space in a unique way should be 3:
$$a' x + b' y + c' z = 1 \tag{7}$$
where $\alpha' = \frac{\alpha}{d}$!
P.S.2. The above method doesn't work if $d = 0$, so one needs at least a boolean variable $d' = 0 \, or \, 1$:
$$a' x + b' y + c' z = d' \tag{8}$$
to represent all possible lines in a 3D space. 
P.S.3. I made a very silly mistake. Eq.1. represents a plane, not a line!

Maybe $5$? $x,y,z$ of the reference point and a unit vector, which needs $\theta$ and $\phi$?
