在边长为1的正三角形$ABC$中，设$\stackrel{\rightharpoonup }{BC}=2\stackrel{\rightharpoonup }{BD},\stackrel{\rightharpoonup }{CA}=3\stackrel{\rightharpoonup }{CE}$，则$\stackrel{\rightharpoonup }{AD}·\stackrel{\rightharpoonup }{BE}=$.