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If $\alpha(s)$ is a unit speed curve with $k\ne0$, how can we show that the equation of the osculating plane through $\alpha(0)$ is $[x-\alpha(0),\alpha'(0),\alpha''(0)] = 0$. (I mean 3 equal bars for the equal sign)

So what I'm thinking is we can use the fact that $[u,v,w] = [u\times v,w]$ and $k = T'/N$, Frenet serret doesn't look too helpful so I'm stuck. The definition of osculating plane is the plane $\alpha(s)$ perpendicular to $B$ (spanned by $T$ and $N$).

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up vote 2 down vote accepted

I suppose that in your question you are using the following notation

\begin{align} &[u,v,w]&&\text{triple product of}\;u,v,w\\ &[u,v]&&\text{inner product of}\;u,v \end{align}

Given that, by definition

\begin{align} T&=\alpha'\\ N&=T'/k=\alpha''/k\\ B&=T\times N \end{align}

I should recall that the equation of a plane of normal $\mathbf{n}=(A,B,C)$ and containing the point $\mathbf{x}_0=(x_0,y_0,z_0)$ is given by

$$ [\mathbf{n},\mathbf{x}-\mathbf{x}_0]=0\implies A(x-x_0)+B(y-y_0)+C(z-z_0)=0 $$

the plane perpendicular to $B$ and passing by $\alpha(0)$ is given by (using your notations)

\begin{align} &[B,\mathbf{x}-\alpha(0)]=0 \\ &[T\times N,\mathbf{x}-\alpha(0)] =0\\ &[T,N,\mathbf{x}-\alpha(0)] =0\\ &[\alpha'(0),\alpha''(0),\mathbf{x}-\alpha(0)]=0 \end{align}

where $k$ has been removed from the last equation, begin not zero.

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I'm not following the notation of the plane P. Can you only use variables which I had in my problem above? Thanks – mary Sep 9 '12 at 23:04
@mary: I tried to be clearer. – enzotib Sep 10 '12 at 8:09
Many Thanks for your help! – mary Sep 10 '12 at 8:35

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