Answer
$r(t) \times r'(t)= \omega a \times b$
Work Step by Step
As we are given that $r(t)=a \cos \omega t+b \sin \omega t$,
Use cross product rule.
$r'(t)=u'(t) \times v(t)+u(t) \times v'(t)$
$r(t)=a \cos \omega t+b \sin \omega t$
and $ r'(t)= -\omega a \sin \omega t+ \omega b \cos \omega t$
Now, $r'(t)=\begin{vmatrix}i&j&k \\\cos \omega t&\sin \omega t&0\\-\omega \sin \omega t&\omega \cos \omega t&0\end{vmatrix}$
This gives: $=\omega c \cos^2 \omega t+ \omega c \sin^2 \omega t$
Now, $\omega a \times b=\begin{vmatrix}a&b&c \\ \omega&0&0\\0&1&1 \end{vmatrix}$
$\omega a \times b=\omega c$
Hence, the result.
$r(t) \times r'(t)= \omega a \times b$
