Question

如图,三角形的外接圆的圆心为0,AB=2,AC=3,BC=根号7,则向量AO×向量BC等于

Answer 1

AO·BC是个定值,与BC都没有关系

令外接圆半径为：R

即：|OA|=|OB|=|OC|=R

AO·BC=AO·(OC-OB)

=OA·OB-OA·OC

=R^2cos-R^2cos

=R^2(2R^2-|AB|^2)/(2R^2)-R^2(2R^2-|AC|^2)/(2R^2)

=R^2-|AB|^2/2-R^2+|AC|^2/2

=|AC|^2/2-|AB|^2/2

=9/2-2=5/2

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或者：取AB边中点D,AC边中点E

则：OD⊥AB,OE⊥AC

故：cos(∠OAB)=|AD|/|AO|=1/R

cos(∠OAC)=|AE|/|AO|=3/(2R)

AO·BC=AO·(AC-AB)

=AO·AC-AO·AB

=R*|AC|cos(∠OAC)-R*|AB|(∠OAB)

=3*R*3/(2R)-R*2/R

=9/2-2

=5/2