Differential Geometry of Curves and Surfaces - Chapter 1 Section 2 Exercise 4

Problem:

Solution:

Consider the function $\alpha(t) \cdot v$, it is a differentiable function of $t$, now $(\alpha(t) \cdot v)' = 2\alpha'(t) \cdot v = 0$ as the vectors are orthogonal. So $\alpha(t) \cdot v$ is a constant function, and therefore $\alpha(t)$ is orthogonal to $v$ for all $t$.