Why is $d(T \times N)/ds$ orthogonal to the unit tangent vector $T$? Consider a curve in $\mathbb{R}^3$ and its Frenet-Serret frame $(T,N,B)$. A calculation gives $$\frac d{ds} (T \times N)= T \times \frac {dN}{ds}.$$
Question:  I read that this implies that $\frac d{ds} (T \times N)$ is orthogonal to $T.$ Why is it so?
 A: Let $\times$ the vector cross product in $\mathbb{R}^3$ and $b,c\in \mathbb{R}^3$, then we have that $a:=b\times c$ is an orthogonal vector to the plane defined by the span of $b$ and $c$, therefore $T\times N'$ is orthogonal to both $T$ and $N':=\frac{\partial}{\partial s}N$.
An elementary proof is given by: let $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$ be vectors in $\mathbb{R}^3$ who coordinates are given by the standard orthonormal basis, then
$$
\langle x\times y, x \rangle=\langle (x_2y_3-x_3y_2,x_3y_1-y_3x_1,x_1y_2-y_1x_2),(x_1,x_2,x_3) \rangle\\
=\color{red}{x_1x_2y_3}\color{blue}{-x_1x_3y_2}+x_2x_3y_1\color{red}{-x_2x_1y_3}\color{blue}{+x_3x_1y_2}-x_3x_2y_1=0
$$
where $\langle \cdot ,\cdot  \rangle$ is the standard dot product in $\mathbb{R}^3$.∎
A: I assume the term "versor" $\equiv$ "vector".
Note that
$\dfrac{d(T \times N)}{ds} = \dot T \times N + T \times \dot N, \tag 1$
where
$\dot T = \dfrac{dT}{ds} \tag 2$
and so forth. We have the Frenet-Serret equations
$\dot T = \kappa N \tag 3$
and
$\dot N = -\kappa T + \tau B; \tag 4$
substituting (3) into (1) yields
$\dfrac{d(T \times N)}{ds} = \kappa N \times N + T \times \dot N = T \times \dot N, \tag 5$
the calculation mentioned in the question itself; we further substitute (4) into (5):
$\dfrac{d(T \times N)}{ds} = T \times(-\kappa T + \tau B)$
$= -\kappa T \times T + \tau T \times B = -\tau N, \tag 6$
and thus
$T \cdot \dfrac{d(T \times N)}{ds} = -\tau T \cdot N = 0. \tag 7$
In short, $d(T \times N)/ds$, being proportional to $N$, is orthogonal to $T$.  $OE\Delta$.
