Curve with all normal planes having a common point

Question

Question:

Consider a regular curve parametrized by arclength. If all of the normal planes have a common point then this curve lies on a sphere.

I guess this has to do with the Frenet frame, but don't know how to proceed...

Answer 1

Let $\gamma: I \to \mathbb R^3$ a curve parameterized by arclength that passes through the fixed point $\gamma (s) \in \mathbb R^3$ in the $n(s)$ direction is given by $$r(\lambda) = \gamma (s) + \lambda n(s)$$

For each $s$ we choose a parameter $\lambda$, which gives us a function $\lambda (s)$. Now let $r_0$ be the intersection point between the curve $\gamma$ and $r(\lambda)$, then we have

$$\gamma(s) = r_0 + \lambda(s)n(s)$$

We simply show that $\lambda$ is constant. Indeed,

$$\begin{align}\lambda (s)^2 &= \lambda(s)n(s) \cdot \lambda(s) n(s) \\&= (\gamma(s) - r_0) (\gamma(s) - r_0)\end{align}$$

Taking the derivative we have

$$\begin{align}(\lambda(s)^2)' &= 2 (\gamma(s) - r_0)' \cdot (\gamma(s) - r_0)\\&=2 \gamma'(s) \lambda (s)n (s)\\&=2 \lambda (s) \underbrace {t(s)\cdot n(s)}_{=0}\\&= 0\end{align}$$ Thus $\lambda(s)^2 = \mathrm{const} \implies \lambda (s) = \mathrm{const}$. Finally

$$\|\gamma(s) - r_0\| = \|\lambda(s)n(s)\| = |\lambda| \|n(s)\| = |\lambda|$$

Therefore $\gamma: I \to S(r_0; |\lambda|)$ where $S(r_0; |\lambda|)$ is the sphere of center $r_0$ and radius $|\lambda|$.

Answer 2

Hint. Suppose that the curve does not lie on a sphere centered on the common point. Then the curve's distance from the common point will not be constant, and therefore, by the Mean Value Theorem, the curve will have a nonzero radial velocity at some parameter value ...

Answer 3

Let $γ=γ(s): I\to\Bbb R^3$ a space curve of unitary velocity. For every $s\in I$ the normal plane of $γ$ is $Π_s:=\{γ(s)+λn(s)+μb(s)|λ,μ\in \Bbb R\}$, and we suppose that there is a fixed point $P$ such that $\forall s \in I:P\in Π_s$. The line $l_s(u)=P+ub(s), u\in \Bbb R$, is contained in $Π_s$ and every vector $γ(s)-l_s(u)$ belongs to $Π_s$, thus it is normal to tangent unitary vector $t(s)$. Consequently $t(s)\cdot (γ(s)-l_s(u))=t(s)\cdot (γ(s)-P+ub(s))$ or $0=t(s)\cdot (γ(s)-P)$ or $\frac {d}{ds}[(γ(s)-P)\cdot (γ(s)-P)]=0$ or $(γ(s)-P)\cdot (γ(s)-P)=constant$ or $||γ(s)-P||=constant$, for every $s\in I$. q.e.d.