How to find a position vector given several other position vectors with a given relationship?

Relative to the origin O , the position vector of A is $6i+6j$ and the position vector of B is $12i-2j$. The point C lies on $\overrightarrow{AB}$− such that $\overrightarrow{AC}=\frac{3}{4}\overrightarrow{AB}$. Find the position vector of C.

As of now, I have found $\overrightarrow{AB}$ and $\overrightarrow{AC}$:

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{AB}=12i-2j-(6i+6j)$

$\overrightarrow{AB}=12i-2j-6i-6j$

$\therefore \overrightarrow{AB}=6i-8j$

$\overrightarrow{AC}=\frac{3}{4}\overrightarrow{AB}$

$\overrightarrow{AC}=\frac{3}{4}\cdot (6i-8j)$

$\overrightarrow{AC}=\frac{3}{4}\cdot 2(3i-4j)$

$\overrightarrow{AC}=\frac{3}{2}\cdot (3i-4j)$

$\overrightarrow{AC}=\frac{3(3i-4j)}{2}$

$\therefore \overrightarrow{AC}=\frac{9i-12j}{2}$

How would $\overrightarrow{OC}$ be found?

Relative to the origin O , the position vector of A is $6i+6j$ and the position vector of B is $12i-2j$. The point C lies on $\overrightarrow{AB}$− such that $\overrightarrow{AC}=\frac{3}{4}\overrightarrow{AB}$. Find the position vector of C.

As of now, I have found $\overrightarrow{AB}$ and $\overrightarrow{AC}$:

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{AB}=12i-2j-(6i+6j)$

$\overrightarrow{AB}=12i-2j-6i-6j$

$\therefore \overrightarrow{AB}=6i-8j$

$\overrightarrow{AC}=\frac{3}{4}\overrightarrow{AB}$

$\overrightarrow{AC}=\frac{3}{4}\cdot (6i-8j)$

$\overrightarrow{AC}=\frac{3}{4}\cdot 2(3i-4j)$

$\overrightarrow{AC}=\frac{3}{2}\cdot (3i-4j)$

$\overrightarrow{AC}=\frac{3(3i-4j)}{2}$

$\therefore \overrightarrow{AC}=\frac{9i-12j}{2}$

How would $\overrightarrow{OC}$ be found?