I have an edge in 3d, which is basically a 3d vector. So, by calculating the normal of the edge, I will have a vector perpendicular to the edge. Therefore, does that mean, if I calculate the normal of that normal, will the original edge vector be returned (since that would be the perpendicular of the perpendicular)? Or, at least, a vector in the correct direction, if not of the same magnitude.
Basically, I'm wondering if normal(normal(vector)) = vector.
If not, why doesn't that work?
The problem is, if you have a vector in $\mathbb{R}^3$ you have infinitely many possible normal vectors. Picture your vector going through the center of a round disk. Then any vector starting in your original vector and "ending" in the disk would be perpendicular. In the same manner, for your normal vector you can find again, infinitely many normal vectors. So, a normal vector to the normal vector of your original vector is not necessarily in the same direction (than your original vector).
You can see this more rigorously in the following way. Let $v=(v_1,v_2,v_3) \in \mathbb{R}^3$ be your original vector. A normal vector would be any $w=(w_1,w_2,w_3) \in \mathbb{R}^3$ such that $v \cdot w = 0$ That is, $v_1w_1 + v_2w_2 + v_3w_3 = 0$. Since you have $1$ equation with $3$ unknowns, (the unknowns are the possible values of your vector $w$). You can easily see there are infinitely many posibilities for $w$. Proceding in the same way, for any $w$ you choose which is orthogonal to $v$, you have again, infinitely many possibilities to choose a $v'$ orthogonal to $w$, which not necessarily implies $v' \cdot v = 0$.
For example, take $v=(1,1,1)$ then taking $w=(-1,1,0)$ we have that $v \cdot w = -1 + 1 + 0 = 0$. Now take $v' = (1,1,0)$ we have that $v' \cdot w = -1 + 1 + 0 = 0 $ However $v' \cdot v = 1 + 1 + 0 = 2$ and therefore $v$ and $v'$ do not have the same direction