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**Problem:**

Let $\gamma: (\alpha,\beta)\to\mathbb{R}^3$ be a regular curve with torsion and curvature that are never $0.$

Show that $\gamma$ is a generalized helix curve if and only if the binormal $b(s)$ makes a fixed angle with a constant vector $a \in \mathbb{R}^3, a\not=0.$

**Definition:**A regular curve $\gamma$ in $\mathbb{R}^3$ with positive curvature is called a generalized helix if its tangent vector makes a fixed angle $\theta$ with a fixed unit vector $a$.

**I know that:**

the curvature $\kappa > 0$

the torsion $\tau >0 $

$n = b \times t$

$t = b \times n$

$b = t \times n$

$t' = \kappa n$

$b' = - \tau n$

**I've tried this so for:**

⟹:

$a$ is constant so $a' =0$. The curvature $k > 0$.

$(a\cdot t)'= a'\cdot t + a\cdot t' = a\cdot t' = a\cdot (\kappa n).$

We know that $\kappa > 0,$ so: $a\cdot n$ must be $0,$ and $n = b \times t$ so $a\cdot(b \times t)$ must be zero - and I'm stuck here.

⟸:

$(a\cdot b)'=a'\cdot b+a\cdot b'=a\cdot b'=-a\cdot (\tau n).$ Since $\tau >0$ then $-a\cdot n$ must be zero (once again i'm stuck).

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