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 $.