The position vector of a particle $P$ is $\vec{r} = (a\cosθ)\vec{i}+ (b\sinθ)\vec{j}+ c\vec{k},$ where $a, b, c$ are constants and $\theta$ is a function of $t$. Given that $\vec{i},\vec{j},\vec{k}$ are unit vectors along right handed rectangular axes $O_x,O_y,O_z$ respectively. Given further that $\dot{\theta}= \omega$ is a constant
(a) Find the components of acceleration in the direction of $OP$ and perpendicular to $OP$, in the case when $a = b$.
$\underline{Attempt}$
Acceleration of $P$ is $\vec{\ddot{r}}=(-a{\omega}^2\cos\theta)\vec{i}+(-a{\omega}^2\sin\theta)\vec{j}$
Now, Acceleration into direction $OP=\dfrac{\vec{\ddot{r}}.\vec{r}}{|\vec{r}|}=\dfrac{((-a{\omega}^2\cos\theta)\vec{i}+(-a{\omega}^2\sin\theta)\vec{j}).(a\cosθ)\vec{i}+ (a\sinθ)\vec{j}+ c\vec{k})}{\sqrt{a^2+c^2}}=\dfrac{-a^2\omega^2}{\sqrt{a^2+c^2}}$
My madam said that was wrong but I don't know what the reason is please give me some hint!!