Showing a parametrized curve is orthogonal 
Let $\alpha : \mathbb{I}\to \mathbb{R}^3$ be a parametrized curve,
  with $\alpha'(t) \ne 0$ for all $t\in I$. Show that $|\alpha(t)|$ is a
  nonzero constant if and only if $\alpha(t)$ is orthogonal to
  $\alpha'(t)$ for all $t\in \mathbb{I}$.

I became really rusty with vector calculus because of a long break (I should have never done that). 
So far I got:
Let,
$\alpha(t) = (x(t), y(t),z(t)),\text{then}$
$\alpha'(t) = (x'(t), y'(t),z'(t))\ne0 $
$|\alpha(t)|= D= \sqrt{x^2+y^2+z^2},\ D'=\frac{xx'+yy'+zz'}{\sqrt{x^2+y^2+z^2}} $
If $\alpha(t)\dot\ \alpha'(t)=0$ then $xx'+yy'+zz'=0$. 
 A: We have $\vert \alpha(t) \vert^2 = \alpha(t) \cdot \alpha(t) = c$, $c$ a constant.  Then
$\frac{d}{dt}(\alpha(t) \cdot \alpha(t)) = 0$,
but
$\frac{d}{dt}(\alpha(t) \cdot \alpha(t)) = \dot{\alpha}(t) \cdot \alpha(t) + \alpha(t) \cdot \dot{\alpha}(t) = 2 \alpha(t) \cdot \dot{\alpha}(t) = 0$,
whence
$\alpha(t) \cdot \dot{\alpha}(t) = 0$.
Sorry about switching up the notation from $\alpha'(t)$ to $\dot{\alpha}(t)$; old dawgs, new tricks; you know how it is! ;)
A: Try this.
Proof
Suppose $|\alpha (t)| \neq 0$, then so does $|\alpha(t)|^2 = \alpha(t) \cdot \alpha(t)$. Differentiating both sides yield $2\alpha'(t)\cdot\alpha(t) = 0 \implies \alpha'(t)\cdot\alpha(t) = 0 $
On the other hand, suppose we have $\alpha'(t)\cdot\alpha(t) = 0$ instead. Consider $|\alpha(t)|^2 = \alpha(t) \cdot \alpha(t)$  again. This time around we have $(|\alpha (t) |^2)' = 2\alpha'(t)\cdot\alpha(t) = 0$ where the RHS of the equation comes from our orthogonal assumption. 
This means that $(|\alpha (t) |^2)' = 0 \implies |\alpha (t) |^2 = c$ for some positive constant $c$. Taking the square root gives the desired result. 
