# A question on the curvature of a regular parametrized curve

The next question is from Do-Carmo's baby book, page 30 question 3 in section 1-6.

The question goes as follows: Show that the curvature $k(t)\neq 0$ of a regular parametrized curve $\alpha : I\rightarrow \mathbb{R}^3$ is the curvature at $t$ of of the plane curve $\pi \circ \alpha$, where $\pi$ is the normal projection of $\alpha$ over the osculating plane at $t$.

Now I guess I don't understand what is $\pi$ here, I mean I have $$\alpha(s)= (x(s),y(s),z(s))$$ assume it's in arclength parametrization, so basically I want to show that:

$$(\pi \circ \alpha (s))'' = k \cdot n_{\pi \circ \alpha (s)}\;,$$

but I am not sure here what $\pi$ equals to, I mean if its a projection on the osculating plane which is the plane of $xy$ then shouldn't it be the binormal to $\alpha$ (in which case it's $(0,0,z(s))$)?

Any hints?

Thanks.

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