Show that $N$ is contained a plane. 
Let $\alpha$ be a planar curve in $\mathbb R^3$, contained in plane $P$. Show that its normal  vector $N$ at every point is in the plane P also.

The only thing I know is that the torsion is 0, but I don't how to relate it to N. Please give me some idea.
 A: Hint: Write down what it means for an arclength-parametrized curve $\alpha$ to be in a plane, and differentiate, using the Frenet equations.
A: Here's how I would do it:
Let the equation of the plane $P$ be given as
$(\mathbf r - \mathbf r_0) \cdot \nu = 0, \tag{1}$
where $\mathbf r_0 \in P$ is a fixed point, $\nu$ is a unit vector normal to $P$, and $\mathbf r \in P$ is an arbitrary point.  Let $\alpha(s)$ be an arc-length parametrized curve lying in $P$; then by (1),
$(\alpha(s) - \mathbf r_0) \cdot \nu = 0, \tag{2}$
and if we differentiate (2) with respect to $s$ we obtain
$\alpha'(s) \cdot \nu = 0 \tag{3}$
since $\mathbf r_0$ and $\nu$ are constants.  Since $\alpha(s)$ is parametrized by arc-length, we have $\alpha'(s) = T(s)$, where $T(s)$ is the unit tangent vector to $\alpha$ and a member of the Frenet frame.  (3) thus becomes
$T(s) \cdot \nu = 0, \tag{4}$
and if we differentiate again and use the Frenet equation $T'(s) = \kappa N(s)$, where $\kappa$ is the curvature of, and $N$ is the unit normal to, the curve $\alpha(s)$, we see that
$\kappa N(s) \cdot \nu = 0, \tag{5}$
assuming of course $T'(s) \ne 0$, which assumption is vital to the definition of $\kappa$ and the Frenet normal $N$.  Under these conditions, $\kappa \ne 0$ and so (5)
becomes 
$N(s) \cdot \nu = 0, \tag{6}$
showing that $N(s)$ is tangent to, i.e., lies in, the plane $P$.  And that's as far as I'm gonna take it.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
