设O中△ABC的外心,
|
三角形的外心是指三边中垂线的交点,延长AO到D,D在BC上AD是BC的中垂线,
AD⊥BC,BD=
BC过点O作OE⊥AB,E在AB上OE是AB的中垂线,AE=1 2
AB1 2
则cos∠BAD=
=AE AO
,AO=AD AB
,AB=|AB•AE AD
|,AE=a
|1 2
|,AD=a
|1 2
+a
|b
∴|
|=AO
=|
|•a
|1 2
|a |
+a
|b 2
,设沿|
|2a |
+a
|b
方向的单位向量为AO
,e
则
=e
+a b |
+a
|b
∴
=|AO
|*AO
=e
•|
|2a |
+a
|b
+a b |
+a
|b
∵
2=|a
|2=|a
|2=b
2b
∴
=AO
(
2a
2+2a
•a
+b
2b
+a
)=b
(
-2a 2(
-2+a
•a
)b
+a
).b
故选A.