According to frenet formulas the unit tangent vector is given as follows.
$ T(t) = \frac{dr/dt}{\lVert r'(t) \rVert}$
And the unit normal vector is given as follows.
$ N(t) = \frac{dT/dt}{\lVert T'(t) \rVert}$
And the binormal is the cross product of those two.
$ B(t) = T(t) \times N(t)$
Here's a visual aid:
I know the binormal is the cross product of those because it needs to perpendicular to both of them. As far as I know the derivate of $N$ will land on the same plane as $B$ and $T$, because of that couldn't $N' = T$ be true? Could anyone help me figure out the zone where $N'$ will land?