VC1.4

Given, $r = \cos nt i + \sin nt j\,$

Since i and j are constant vectors, $\frac{di}{dt}=\frac{dj}{dk}=0\,$

Therefore,from the given $\frac{dr}{dt}=-n\sin nt i+n\cos nt j \,$

$r\times \frac{dr}{dt}=(\cos nt i + \sin nt j)\times (-n\sin nt i+n\cos nt j)\,$

$-n[-\cos^2 nt i\times j + \sin^2 nt j\times i]\,$

$-n[-cos^2 nt i\times j - \sin^2 nt i\times j]\,$

$n[\cos^2 nt + \sin^2 nt]i\times j = nk\,$ (Since $i\times j = k\,$

And $r\cdot (\frac{dr}{dt})=(\cos nt i + \sin nt j)[-n(\sin nt i - \cos nt j)]=-n(\cos nt \sin nt - \sin nt \cos nt) = 0\,$

Hence proved.