binormal vectors and generalized helix There's a problem I'm stucked on it. I wonder if anybody is able to help me because I tried almost every idea that I could thougth of. Here it is:
If $ u $ is a fixed direction and for any point $ s $ of a space curve
$ < B(s) , u > = constant $
holds then prove that there's a constant vector $ v $ such that for every point $ s $ we have
$ < T,v > = constant $
I wish somebody could help me.
[ $ B $ is the binormal vector of curve. ]
[ the curve is prameterized with arc length ]

If $\langle B(s), u\rangle = c$ then $\langle B'(s), u\rangle + \langle B(s), u'\rangle = 0$. But $u$ is constant, so $\langle B(s), u'\rangle = 0$ and therefore $\langle B(s), u\rangle ' = \langle B'(s), u\rangle = 0$.
Using the Frenet Serret formulas, $\langle B'(s), u\rangle = \tau(s) \langle N(s), u\rangle = 0$ , so either the torsion is zero or $\langle N(s), u\rangle$ is zero.
Let $\alpha(s)$ be the curve such that its binormal vector makes a constant angle with $u$. 


*

*Case 1: If $\tau(s)$ is thero then the curve lies in its osculating plane and the binormal vector ( which is the normal vector of that plane ) is constant, so $T$ makes a constant angle with the constant direction $B$.

*Case 2: if $\langle N(s), u\rangle=0$ then $\langle N(s), u\rangle ' = 0$, so $\langle N'(s), u\rangle = 0$. On the other hand, $\langle T(s), u\rangle ' = \langle T'(s), u\rangle $ and using the Frenet Serret formulas again $\langle T'(s), u\rangle  = k(s)\langle N(s), u\rangle = k(s) \cdot 0 = 0$, so $\langle T(s), u\rangle $ is constant.
