The position vector of the center of mass $$\vec{r_{cm}}$$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

A

$$\vec{r_{cm}} = \frac{13}{8}L\hat{x} + \frac{5}{8}L\hat{y}$$

B

$$\vec{r_{cm}} = \frac{5}{8}L\hat{x} + \frac{13}{8}L\hat{y}$$

C

$$\vec{r_{cm}} = \frac{3}{8}L\hat{x} + \frac{11}{8}L\hat{y}$$

D

$$\vec{r_{cm}} = \frac{11}{8}L\hat{x} + \frac{3}{8}L\hat{y}$$