点O是锐角三角形ABC的外心,AB=8,AC=12,∠A=60°若向量AO=x向量AB+y向量AC,则2x+3y=

Question

点O是锐角三角形ABC的外心,AB=8,AC=12,∠A=60°若向量AO=x向量AB+y向量AC,则2x+3y=

Answer 1

AO*AB=1/2*|AB|^2 =8^2/2=32 AB*AC=|AB||AC|cosA=8*12*cos60°=48
所以 AO*AB=(xAB+yAC)*AB=x*AB^2+yAB*AC=64x+48y=32===>4x+3y=2 .(1).
同理 AO*AC=1/2*|AC|^2=12^2/2=72
AO*AC=(xAB+yAC)*AC=x*AB*AC+yAC*AC=48x+144y=72===>2x+6y=3.(2)
(1)+(2),得 6x+9y=5 ===> 2x+3y=5/3.