A particle moves so that its position vector is given by r→=cosωt x^+sinωt y^, where ω is a constant.Which of the following is true?

Given, r→=cosωt x^+sinωt y^∴v→=dr→dt=-ωsinωt x^+ωcosωt y^a→=dv→dt=-ω2cosωt x^-ω2sinωt y^=-ω2r→ Since position vector (r→) is directed away from the origin, so,  acceleration -ω2r→ is directed towards the origin. also,⇒r→⊥v→r→·v→=(cosωtx^+sinωty^)·(-ωsinωtx^+ωcosωty^)=-ωsinωtcosωt+ωsinωtcosωt=0