The Frenet frame formula says that the first derivation of the equation $q(t)$ is my view:
$$q'(t) = \verb|vec_view|$$
the cross product of derivation one and two $q' \times q''$ is my binominal vector
In some formulary, equations are only derived like $$\verb|vec_view| = q',$$ $$\verb|Binominal vector| = q' \times q''.$$
In some other their are divided through the normal vector
$$\verb|Binominal vector| = \frac{q'}{\Vert q'\Vert} \times \frac{q''}{\Vert q''\Vert}.$$
I am not sure why some formularies are not dividing. In my opinion the formulary with $$\verb|vec_view| = \frac{q'}{\Vert q'\Vert}$$ is the right one, because I'll get a normalized vector.
Am I right ?
I don't know the programming background to your question; it's possible that in whatever applications are being considered, normalization of all the Frénet-Serret vectors isn't needed, so that as long as $q$ is already normalized with $\|\dot{q}\|=\|T\|=1$ (i.e. $q$ has natural parametrization), one can observe that $\ddot{q}$ is parallel to the normal vector $N=\dot{T}/\|\dot{T}\|=\ddot{q}/\|\ddot{q}\|$ and thus $\dot{q}\times\ddot{q}$ is parallel to the binormal vector $B=T\times N$. However, these do not really make up the Frénet-Serret moving frame according to its definition, but rather an augmented variation that might be in place because of outside considerations. You are correct that the true Frénet-Serret frame, by definition, is fully normalized, $\|T\|=\|N\|=\|B\|=1$. If normalization is, in fact, needed for whatever purposes are at hand in your context, then there will be potentially serious problems with failing to normalize. Moreover, if $q$ isn't naturally parametrized (according to its arclength), then the resulting vectors won't even be parallel to the vectors from the true frame, making even bigger problems.
(Also, to nitpick: it's "binormal" instead of "binominal," and I don't even think "formulary" is a real word in mathematics - it's a pharmaceutical list - though perhaps you're just being kooky on the latter term.)