Showing a parametrized curve is orthogonal

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 $xx'+yy'+zz'=0$.