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Given,r=\cos nti+\sin ntj\,

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

Therefore,from the given {\frac  {dr}{dt}}=-n\sin nti+n\cos ntj\,

r\times {\frac  {dr}{dt}}=(\cos nti+\sin ntj)\times (-n\sin nti+n\cos ntj)\,

-n[-\cos ^{2}nti\times j+\sin ^{2}ntj\times i]\,

-n[-cos^{2}nti\times j-\sin ^{2}nti\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 nti+\sin ntj)[-n(\sin nti-\cos ntj)]=-n(\cos nt\sin nt-\sin nt\cos nt)=0\,

Hence proved.

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