Generalized helix curve Problem:
Let $ γ(α,β): →\Bbb R^3 $ be a regular curve with torsion and curvature that are never zero. 
Show that $γ$ is a generalized helix curve if and only if binormal $b(s)$ has a fixed angle with a constant vector $a$ in $\Bbb R^3, a≠0 $  
Definition:  A regular curve $γ$ in $ℝ^3$ with curvature $> 0$ is called a generalized helix if its tangent vector makes a fixed angle $\theta$ with a fixed unit vector $a$. 
This is what I've tried so far:
⟹:
$a$ is constant so $a'=0$
The curvature $\kappa > 0$
$ 0 = (a\cdot t)'= a't + at' = at' = a\kappa n$ 
we know that $\kappa > 0$, so:
$a\cdot n = 0$  and
and $n = b x t$
so $a·(b x t) = 0$ 
(From here I'm stuck)
⟸:
$0 =(a·b)'=a'·b+a·b'=a·b'=-a·\tau n$
since $\tau >0$ then $-a·n =0$ 
(Once again I'm stuck)
 A: Let $a$ be our fixed vector; assuming $\gamma(s)$ is parametrized by the arc-length $s$, which is always possible for a regular curve, we have the unit tangent vector $T$:
$T(s) = \gamma^\prime(s); \tag 1$
since $T(s)$ makes a constant angle $\theta$ with $a$, we have
$T(s) \cdot a = \vert T(s) \vert \vert a \vert \cos \theta = \vert a \vert \cos \theta, \tag 2$
since $\vert T(s) \vert = 1$; if we set
$\alpha = \dfrac{a}{\vert a \vert}, \tag 3$
then $\alpha$ is easily seen to be as unit vector and (2) yields
$T(s) \cdot \alpha = T(s) \cdot \dfrac{a}{\vert a \vert} = \cos \theta; \tag 4$
we may differentiate (4) to obtain
$T^\prime(s) \cdot \alpha = (\cos \theta)^\prime = 0, \tag 5$
since $\cos \theta$ and $\alpha$ are both constant; by the Frenet-Serret equation
$T'(s) = \kappa(s) N(s), \tag  6$
where $N(s)$ is the unit normal to $\gamma(s)$, we thus find
$\kappa(s) N(s) \cdot \alpha = 0, \tag 7$
and since by hytothesis $\kappa(s) \ne 0$ we infer that
$N(s) \cdot \alpha = 0; \tag 8$
next, letting $\phi(s)$ be the angle 'twixt $\alpha$ and the binormal to $\gamma(s)$, $B(s)$, we have
$\alpha \cdot B(s) = \vert \alpha \vert \vert B(s) \vert \cos \phi(s) = \cos \phi(s), \tag 9$
since $\vert B(s) \vert = 1$; (9) yields
$(\cos \phi(s))' = (\alpha \cdot B(s))' = \alpha \cdot B'(s); \tag{10}$
again invoking Frenet-Serret, we have
$B'(s) = -\tau(s) N(s); \tag{11}$
inserting this in (10) and also using (8) we find
$(\cos \phi(s))' = (\alpha \cdot B(s))' = \alpha \cdot (-\tau(s) N(s)) = -\tau(s) \alpha \cdot N(s) = 0; \tag{12}$
since $\cos \phi(s)$ is constant, $\phi(s)$ is constant as well.  We have thus shown that the binormal vector of a generalized helix makes a constant angle with the unit vector $\alpha$.
The above argument may be reversed; if $\phi(s)$ is constant and $\tau(s) \ne 0$, then (10)-(11) show that (8) binds; from there, we need merely walk the steps backward to see that $\cos \theta(s)$, and hence $\theta(s)$, is constant.
Note Added in Edit, Wednesday 28 February 2018 8:54 PM PST:  Consider equation (4); by virtue of (1), it may be written
$\gamma'(s) \cdot \alpha = \cos \theta; \tag{13}$
thus, for any $s_0, s$ in the interval of definition of $\gamma(s)$ we have
$\gamma(s) \cdot \alpha - \gamma(s_0) \cdot \alpha = \displaystyle \int_{s_0}^s \gamma'(u) \cdot \alpha \; du$
$= \displaystyle \int_{s_0}^s T(u) \cdot \alpha \; du = \int_{s_0}^s \cos \theta \; du = (\cos \theta) (s - s_0); \tag{14}$
this indicates that the component of the vector $\gamma(s)$ along an axis determined by $\alpha$, in the $\alpha$ direction, is in fact linear in $s$.  If we now construct the plane $P$ normal to $\alpha$, we can describe $\gamma(s)$ in entirety by specifying two more coordinates in this plane.  It would be informative, I'll warrant, to write out the equations for the Frenet-Serret framing of $\gamma(s)$ in such coordinates.  Just as (4) expresses the projection of $T(s) = \gamma'(s)$ onto the $\alpha$-axis, so we would obtain corresponding equations for the projection of $\gamma(s)$ onto $P$.  My curiosity is piqued.  End of Note.
