# Admissible curve contained in a plane

(Assume a unit speed parametrization)

Prove that an admissible curve $c:(a,b) \to \mathbb R^3$ (hence $c^{\prime} \times c^{\prime\prime}$ is never zero) with a zero torsion is contained in a plane.

I understand why this should be true, if there is no torsion element to the curve then $c$ can't twist into the third dimension.

$\mathbf T = c^{\prime}$

$\mathbf N = c^{\prime\prime}$

$\mathbf B = \mathbf T \times \mathbf N$

$\mathbf T^{\prime} = \kappa \mathbf N$

$\mathbf N^{\prime} = -\kappa \mathbf T + \tau\mathbf B$

$\mathbf B^{\prime} = -\tau\mathbf N$

So if the torsion $\tau$ is zero, $\mathbf B^{\prime} = \mathbf 0$, therefore $\mathbf B$ is a constant not equal to zero. I'm not sure where to go from here. If the normal vector of $\mathbf T$ and $\mathbf N$ is a constant, does that mean no matter what point you are at on the curve $\mathbf B$ points in the same direction?

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"If the binormal vector is a constant, does that mean no matter what point you are at on the curve B points in the same direction?" - yes. –  Ｊ. Ｍ. Dec 3 '10 at 0:03
$T=c'$ means that $c$ is assumed unit speed, but $N=c''$ is incorrect. If $c$ is unit speed, then $c''=T'=\kappa N$. –  Jonas Meyer Dec 3 '10 at 2:36
Show that if $t_0$ is in $(a,b)$, then for all $t\in(a,b)$, $c(t)-c(t_0)$ lies in the plane through the origin orthogonal to (the constant vector) $B$ by showing that the function $f(t)=(c(t)-c(t_0))\cdot B$ is identically zero.