Ey 12 E B A4兀5r E 1 D E EAE EB=4兀 3 AX 结论 E3 0 2 4 1 r p EA     = 结论 3 1 r E  3 4 0 1 r p EB     = − l r y x • • B A l  r E+  E+  E−  E−  EB  EA  E p  E  p  E  p  E  p  