Answer
$r(t) \times r'(t)= \omega a \times b$.
Work Step by Step
Given: $r(t)=a \cos \omega t+b \sin \omega t$,
Our aim is to show that $r(t) \times r'(t)= \omega a \times b$.
In order to find this, we will use the cross product rule.
$r'(t)=u'(t) \times v(t)+u(t) \times v'(t)$
$r(t)=a \cos \omega t+b \sin \omega t \implies r'(t)= -\omega a \sin \omega t+ \omega b \cos \omega t$
From the question, we take left side: $r(t) \times r'(t)=\begin{vmatrix}i&j&k \\\cos \omega t&\sin \omega t&0\\-\omega \sin \omega t&\omega \cos \omega t&0\end{vmatrix}=\omega c \cos^2 \omega t+ \omega c \sin^2 \omega t$
Since,$sin^2t+cos^2t=1$
Thus,
$r(t) \times r'(t)=wc$ ....(1)
From the question, we take right side:$\omega a \times b=\begin{vmatrix}a&b&c \\ \omega&0&0\\0&1&1 \end{vmatrix}$
$\omega a \times b=\omega c$ ...(2)
From equations (1) and (2), we have
$r(t) \times r'(t)= \omega a \times b$.
