If \(\vec{L}\) and \(\vec{P}\) represent the angular momentum and linear momentum respectively of a particle of mass ' \…

Given position vector of particle in circular motion.
\(\vec{r}=a\cos \omega t\overset{^}{i}+a\sin \omega t\overset{^}{j}\)
Velocity is time derivative of position.
\(\vec{v}=\frac{d\vec{r}}{\mathrm{dt}}=−a\omega \sin \omega t\overset{^}{i}+a\omega \cos \omega t\overset{^}{j}\)
Acceleration is time derivative of velocity.
\(\vec{a}=\frac{d\vec{v}}{\mathrm{dt}}=−a{\omega }^{2}\cos \omega t\overset{^}{i}−a{\omega }^{2}\sin \omega t\overset{^}{j}\)
Force is proportional to acceleration.
\(\vec{F}=m\vec{a}\)
Acceleration vector is opposite in direction to position vector.
Therefore acceleration is antiparallel to position vector.