Let $\vec{\alpha}:I\to \mathbb{R}^3$ be an arclength parametrized curve such that $\kappa(s)>0$ and $\tau(s)\neq 0$ for all $s \in I.$ If $\vec{\alpha}''(s)\cdot \vec{u}=0$ for all $s\in I$ for some $\vec{u}\in \mathbb{R}^3$, prove that $$\tau(s)+\left(\frac{\kappa(s)}{\tau(s)}\right)'=0.$$
Attempt. One thought is that since $\vec{\alpha}''(s)\cdot \vec{u}=0$ for all $s\in I$, we get $\vec{\alpha}'\cdot \vec{u}$ is constant in $I$ and the curve is a generalized helix. In that case $\frac{\kappa}{\tau}$ is constant and the desired equality becomes $\tau =0$, which contradicts the given hypothesis about the torsion. Is there a flaw in the above argument?
Another thought was that curve $\vec{\alpha}''$ is planar, so it's torsion is zero, i.e. $(\vec{\alpha}''',\vec{\alpha}'''',\vec{\alpha}''''')=0.$
Thank in advance for the help.
I think your argument is correct and that the statement is flawed. One has: $\kappa(s)/\tau(s)$ is constant. Here are my calculations:
We may assume that $u=(0,0,1)$ so that $n(s)=(\cos \phi(s),\sin \phi(s),0)$, for some function $\phi(s)$. Now, $t(s)\cdot n(s)=0$, so $t(s)=(-\alpha \sin \phi(s), \alpha \cos \phi(s), \beta)$ with $\alpha^2+\beta^2=1$. A priory $\alpha$ and $\beta$ depends upon $s$ but here is where the auxiliary condition enters: $t'(s)\cdot u=0$ so $\beta$ is a constant, whence also $\alpha$.
Now $b=t\times n= (\beta \sin \phi(s), -\beta \cos \phi(s), \alpha)$, and then taking derivatives you verify: $t'(s) = \kappa(s) n(s)$ with $\kappa(s)=-\alpha \phi'(s)$ and $b'(s)=-\tau(s) \phi'(s)$ with $\tau(s) = -\beta \phi'(s)$. In other words, $\kappa(s)/\tau(s) = \alpha/\beta$ is a constant.