Prove that the curve is on a circle

Question

Consider a curve $\alpha: I\rightarrow \mathbb{R}^3$ parameterized by arc length such that all the normals contain a comum point. Prove that the curve is on a circle.

Well, we can suppose that this point is the zero. So, there is a function $\lambda$ s.t. $\alpha(s)=\lambda(s)n(s)\quad \forall s\in I$... Right?

The book recommends derive this equation:

$$\alpha'(s)=\lambda'(s)n(s)+\lambda(s)n'(s)$$

$$t(s)=\lambda'(s)n(s)-\lambda(s)k(s)t(s)-\tau(s)b(s)\text{ (Frenet's formula)}$$

$$t(s)(1+\lambda(s)k(s))-\lambda'(s)n(s)+\tau(s)b(s)=0$$

How can I conclude of here that $\tau=0$ and $\lambda=ctc.$?

Many thanks.

Answer 1

Your last equation is a vector equation. The standard way to extract information from vector equations is to take dot (and cross) products with vectors significant in the equation. In this case, $t(s),n(s),b(s)$ are orthogonal for each $s$, so

• dotting with $b(s)$ gives $0 + 0 + \tau(s) b(s) \cdot b(s) = 0 $ so $\tau(s)=0$,
• dotting with $n(s)$ gives $ 0 + \lambda'(s) n(s) \cdot n(s) + 0 = 0 $ so $\lambda'(s) = 0$ and $\lambda(s)$,
• dotting with $t(s)$ tells you about the curvature's relation to $\lambda$, namely $k(s) = -1/\lambda(s)$.