Let $a:I \to \mathbb{R}^3$ be a normal curve with unit speed. Denote $\kappa(t), \tau(t)$ the curvature and torsion of $a$ and assume that $\kappa(t) > 0$ and $\tau(t)\neq 0$ for all $t\in I$ and that there exists a constant vector $v$ satisfying $\langle a''(t), v\rangle = 0$ for all $t\in I$. Show that $$\tau(t) + \left( \frac{\kappa(t)}{\tau(t)}\right)^{'} = 0 \quad \forall\; t\in I$$
Attempt: My idea here is to use Frenet-Serret equations. $\langle a''(t), v\rangle = 0$ implies that $\langle N, v \rangle = 0$. From here neither writing $$v = \langle T, v\rangle T + \langle B, v \rangle B$$ nor differentiating the initial equality led to something meaningful. Another idea was to note that $N(t)$ is actually a planar curve so it must have torsion equal to zero but the calculations led to nothing.
Any help is appreciated!