Answer
True
Work Step by Step
Given; $|r(t)|=1$ for all the values of $t$
$|r(t)|$ can be written as;
$|r(t)|=r(t).r(t)$
Thus,
$r(t).r(t)=1$
Let us find the derivative.
$0=|r(t)|=r'(t).r(t)+r(t).r'(t)$
$2r'(t).r(t)=0$
This implies that
$r'(t).r(t)=0$
Therefore, $r'(t)$ is orthogonal to $r(t)$ for any value of $t$.
Hence, the given statement is true.
