设D为BC中点,则AD=(AB+AC)/2
点O为△ABC的外心,故OB=OC,
又OD为等腰△OBC中线,故OD与BC垂直, 向量OD•BC=0
于是
AO•BC
=(AD+DO)•BC
=AD•BC+DO•BC
=[(AB+AC)/2]•BC+0
=[(AB+AC)/2]•(AC-AB)
=(AC•AC-AB•AB)/2
=(|AC|^2-|AB|^2)/2
=(4^2-2^2)/2
=6
设D为BC中点,则AD=(AB+AC)/2
点O为△ABC的外心,故OB=OC,
又OD为等腰△OBC中线,故OD与BC垂直, 向量OD•BC=0
于是
AO•BC
=(AD+DO)•BC
=AD•BC+DO•BC
=[(AB+AC)/2]•BC+0
=[(AB+AC)/2]•(AC-AB)
=(AC•AC-AB•AB)/2
=(|AC|^2-|AB|^2)/2
=(4^2-2^2)/2
=6