由题意知$$\textbar\vec{a}-\vec{b}{{\textbar}^{2}}={{(\cos \alpha -\cos \beta )}^{2}}+{{(\sin \alpha -\sin \beta )}^{2}}$$  $$=2-2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )=2-2\cos (\alpha -\beta )$$，  则$$2-2\cos (\alpha -\beta )=\frac{4}{5}$$，$$\cos (\alpha -\beta )=\frac{3}{5}$$．  由$$\alpha =(\alpha -\beta )+\beta $$，且$$0\textless{}\alpha \textless{}\frac{ \pi }{2}$$，$$-\frac{ \pi }{2}\textless{}\beta \textless{}0$$及$$\sin \beta =-\frac{5}{13}$$，$$\cos (\alpha -\beta )=\frac{3}{5}$$，  知$$\cos \beta =\frac{12}{13}$$，$$\sin (\alpha -\beta )=\frac{4}{5}$$，  从而得$$\sin \alpha =\sin [\beta +(\alpha -\beta )]=\sin \beta \cos (\alpha -\beta )+\cos \beta \sin (\alpha -\beta )$$  $$=-\frac{5}{13}\times \frac{3}{5}+\frac{12}{13}\times \frac{4}{5}=\frac{33}{65}$$，  故选$$\text{C}$$． 
已知$$\vec{a}=(\cos \alpha ,\sin \alpha )$$，$$\vec{b}=(\cos \beta ,\sin \beta )$$，$$\textbar\vec{a}-\vec{b}\textbar=\frac{2\sqrt{5}}{5}$$，若$$0\textless{}\alpha \textless{}\frac{ \pi }{2}$$，$$-\frac{ \pi }{2}\textless{}\beta \textless{}0$$，且$$\sin \beta =-\frac{5}{13}$$，则$$\sin \alpha =$$．
Options:
- $$\frac{31}{65}$$ 
- $$\frac{32}{65}$$ 
- $$\frac{33}{65}$$ 
- $$\frac{34}{65}$$ 