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Let $\alpha$ be a curve such that $|\alpha'(s)|=1$ for all $t$ and $k\neq 0$. The tangent vector $\vec T(s)$ and the normal vector $\vec N(s)$ through $\alpha(s)$ span a plane called the osculating plane. Show that $\alpha$ is a plane curve if and only if there exists a point $\vec x_0\in\mathbb R^3$ such that every osculating plane spanned by $\vec T$ and $\vec N$ contains $\vec x_0$.

So I started showing the "$\Rightarrow$" by assuming there exists $c_1,c_2$ such that $\vec x_0 = \alpha(s) + c_1\vec T + c_2\vec N$ and tried to find the values of $c_1,c_2$ by differentiating and using the Frenet-Serret-equations, but this didn't really get me anywhere.

Can anyone help me with this? And how do I show the "$\Leftarrow$" part? Thanks in advance!

wchargin
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dinosaur
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    Hint: consider the torsion given by $$\tau=\frac{\left(\alpha'\times \alpha''\right)\cdot \alpha'''}{\left|\left|\alpha'\times \alpha''\right|\right|^2}$$ – wchargin Oct 25 '13 at 17:22
  • I don't really know where to go with this. I know that if $\alpha$ is a plane curve the torsion $\tau=0$. So considering my assumption I differentiated and got $0=\vec T + c_1\vec T' + c_2\vec N' = \vec T + c_1k\vec N - c_2k\vec T$, which led to $c_1=0$ and $c_2=\frac 1 k$. So every plane $E:\vec x = \alpha(s)+\lambda \vec T + \mu\vec N$ contains the point $x_0=\alpha(s)+\frac 1 k \vec N$, which is the $\Rightarrow$-part. Is that correct? – dinosaur Oct 25 '13 at 18:00
  • @WChargin Yes, remembering that the nominator equals $\det(\alpha',\alpha'',\alpha''')$. – Michael Hoppe Oct 25 '13 at 19:12
  • okay, so I think I also got the $\Leftarrow$-part: Write $\alpha(s)=\vec x_0+c_1\vec T(s)+c_2\vec N(s)$ and by differentiating we get $\alpha'(s)=c_1\vec T'+c_2\vec N' \Leftrightarrow 0=(-1-c_2k)\vec T + c_1k\vec N +c_2\tau\vec B$. Comparing coefficients we get $c_1=0$ since $k\neq 0$ and $c_2=-\frac 1 k$. We also get $\tau\neq 0$ since $c_2\neq 0$. Since $\tau=0$ $\alpha$ is a plane curve. – dinosaur Oct 26 '13 at 09:41
  • You are assuming that the coefficients $c_1$ and $c_2$ are constant when they are functions. (For each $s$, we may need different coefficients) – Guillermo Mosse Oct 05 '16 at 15:02

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