Let $\displaystyle \alpha\colon I\rightarrow R^3$ be a parametrized curve and let $\displaystyle \mathbf{v} \in R^3$ be a fixed vector. Assume that $\displaystyle \alpha'(t)$ is orthogonal to $\displaystyle \mathbf{v}$

for all $\displaystyle t \in I$ and that $\displaystyle \alpha(0)$ is also orthogonal to $\displaystyle \mathbf{v}$

Prove that $\displaystyle \alpha(t)$ is orthogonal to $\displaystyle \mathbf{v}$ for all $\displaystyle t \in I$

We are given that $\displaystyle \alpha'(t)$ and $\displaystyle \alpha(0)$ is orthogonal to $\displaystyle \mathbf{v}$

Let $\displaystyle \mathbf{v} = (v_1, v_2, v_3)$ and $\displaystyle \alpha(t) = (x(t), y(t), z(t))$

$\displaystyle \therefore \alpha'(t) = (x'(t), y'(t), z'(t)) $

So we have:

$\displaystyle \alpha'(t) \cdot\, \mathbf{v} = (x'(t) * \, v_1 +\,\, y'(t) * v_2 + \,\, z'(t) * v_3) = 0 \text{ [Given]}$

$\displaystyle \alpha(0) \cdot\, \mathbf{v} = (x(0) *\, v_1 +\,\, y(0) * v_2 +\,\, z(0) * v_3) = 0 \text{ [Given]}$

Now how do I use this info and prove that $\displaystyle \alpha(t)$ is orthogonal to $\displaystyle \mathbf{v}$ for all $\displaystyle t \in I$?

Is it possible for anyone to kindly give some hints on how to prove this?