I was trying to solve the following exercise from Shifrin's textbook (ex. 9 of section 1.2)
Prove that if all the normal planes of a curve pass through a particular point, then the curve lies on a sphere
My attempt was the following: consider $\alpha:I \longrightarrow \mathbb{R}^3$ an arclength-parametrized curve. The normal plane at point $s \in I$ is the plane $\langle N(s),B(s) \rangle$. From the hypotesis we have that $\exists P \in \mathbb{R}^3$ such that for every $s$ we have that $$\langle N(s),B(s) \rangle =\{x \in \mathbb{R}^3 \mid T(s) \cdot (x-P)=0\} $$ So I wanted to show that $\alpha (s) \in \langle N(s),B(s) \rangle$, because this implies that $\alpha '(s) \cdot (\alpha (s)-P)=0$, but $2\alpha '(s) \cdot (\alpha (s)-P)=f'(s)$ where $f(s)=(\alpha (s)-P) \cdot (\alpha (s)-P)=\parallel \alpha(s)-P \parallel ^2$, and so we can conclude that $\alpha$ lies on a sphere because its norm is constant.
The problem is that I don't know how to show that $\alpha(s) \in \langle N(s),B(s) \rangle$ and the second answer to this question also made me think if the normal plane is $\langle N(s),B(s) \rangle$ or the translated $\alpha(s) + \langle N(s),B(s) \rangle$.
So my question is: what is the exact definition of the normal plane? And is my attempt correct? How to conclude it?
Edit: maybe I figured out how to solve it. Since, from what I understood, the normal plane is the plane spanned by $N(s),B(s)$ that pass through $\alpha(s)$, we can describe it with $$\Pi _s=\{x \in \mathbb{R}^3 \mid T(s) \cdot (x-\alpha(s))=0\}$$ Since we know from hypotesis that $P \in \Pi _s$ we have that $T(s) \cdot (P-\alpha(s))=0$ for every $s$. Thus we can conclude as I said before. Is it correct?
And another question: does $\Pi _s = \alpha(s) + \langle N(s),B(s) \rangle$?
We assume that all the normal planes to the curve $\alpha(s)$ pass through a the origin $0$ of the euclidian space.
As $\alpha'(s)$ is tangent top the curve, it perpendicular to $\alpha (s)$. Then $<\alpha (s). \alpha(s)>=F(s)$ satisfies $F'(s)= 2<\alpha (s) . \alpha'(s)>=0$. In other words the square of the distance of $\alpha (s)$ to the origin is constant. And the curve lies on a sphere.
If the special point is not the origin, just change the origin.