Proving that two curves in $\mathbb{R^3}$ with the same binormal vector are congruent 
Let $\alpha, \bar{\alpha}: I \mapsto \mathbb{R^3}$ be two regular unit speed curves with non vanishing curvature and torsion. Prove that if the binormal vectors of the curves coincide, i.e $B(s) = \bar{B}(s)$, they are congruent.

I know there is a unique isometry that takes the orthonormal frenet frame of $\alpha$ to that of $\bar{\alpha}$, but I don't know how to prove what the exercise is asking me. I  tried a proof by contradiction but that led nowhere. I'd be grateful for any help.
 A: If $B(s) = \overline{B}(s)$, differentiation gives $\tau(s)N(s) = \overline{\tau}(s)\overline{N}(s)$. Since torsions are positive and normals are unit vectors, applying $|\cdot|$ says that $|\tau(s)| = |\overline{\tau}(s)|$, and hence $N(s) = \pm\overline{N}(s)$. Reparametrizing one of the curves we can assume that $N(s) = \overline{N}(s)$. Differentiating again says $$-\kappa(s) T(s) + \tau(s)B(s) = -\overline{\kappa}(s)\overline{T}(s) + \overline{\tau}(s)\overline{B}(s),$$and by previous work we have that $\kappa(s)T(s)  = \overline{\kappa}(s)\overline{T}(s)$. Similarly to what we have done to the torsion, we have $\kappa(s) = \overline{\kappa}(s)$. Since $\kappa = \overline{\kappa}$ and $\tau = \overline{\tau}$, we're done.
A: We have
$$B'(s) = \tau (s) N(s) \qquad \text{and} \qquad \overline{B}'(s) = \overline{\tau} (s) \overline{N}(s)$$
since
$$B(s) = \overline{B}(s), \quad B'(s) = \overline{B}'(s)$$
then
$$\tau(s) N(s) = \overline{\tau}(s) \overline{N}(s).$$
We have that $\tau(s) \neq 0$ so
$$N(s) = \frac{\overline{\tau}(s)}{\tau(s)} \cdot \overline{N}(s).$$
Taking the norm
$$\|N(s)\| = \left| \frac{\overline{\tau}(s)}{\tau(s)}\right| \cdot \|\overline{N}(s)\| \Rightarrow \left| \frac{\overline{\tau}(s)}{\tau(s)}\right| =1 \Rightarrow |\overline{\tau}(s)| = |\tau(s)|$$
then, $N(s) = \pm \overline{N}(s)$. So
$$\frac{\alpha''(s)}{\kappa(s)} = \pm \frac{\alpha''(s)}{\overline{\kappa(s)}} \Rightarrow \frac{\|\alpha''(s)\|}{\kappa(s)} = \frac{\|\alpha''(s)\|}{\overline{\kappa}(s)} = 1$$
$$\Rightarrow \frac{\overline{\kappa(s)}}{\kappa(s)}=1 \Rightarrow \overline{\kappa}(s) = \kappa(s).$$
As the curves are equal and the torsions coincide at less than sign, so by the fundamental theorem of the curves, there is an isometry $F: \mathbb{R}^3 \to \mathbb{R}^3$ such that $\overline{\alpha} = \alpha \circ F$. Therefore, $\alpha$ and $\overline{\alpha}$ are congruents.
