Question

已知

OA

+

OB

+

OC

=

0

，

OA

•

OB

=

OB

•

OC

=

OC

•

OA

=-1．
（1）求|

OA

|；
（2）試判斷△ABC的形狀，并求其面積．

Answer 1

考點：三角形的形狀判斷,平面向量數量積的運算

專題：解三角形

分析：（1）由題意可得

OA

=-

OB

-

OC

，平方結合已知可得

OB

2+

OC

2=

OA

2+2，由

OA

+

OB

+

OC

=

0

兩邊平方可得

OA

2+

OB

2+

OC

2-6=0，兩式綜合可得

OA

2=4，進而可得|

OA

|=2；
（2）同理可得|

OB

|=|

OC

|=|

OA

|=2，即△ABC為正三角形，易得面積．

解答： 解：（1）∵

OA

+

OB

+

OC

=

0

，∴

OA

=-

OB

-

OC

，
平方可得

OA

2=（

OB

+

OC

）²=

OB

2+

OC

2+2

OB

•

OC

=

OB

2+

OC

2-2，
∴

OB

2+

OC

2=

OA

2+2，①
由

OA

+

OB

+

OC

=

0

兩邊平方可得

OA

2+

OB

2+

OC

2+2

OA

•

OB

+2

OB

•

OC

+2

OC

•

OA

=0，
∴

OA

2+

OB

2+

OC

2-6=0，②
把①代入②可得

OA

2+

OA

2+2-6=0，即

OA

2=4，∴|

OA

|=2，
（2）同理可得|

OB

|=|

OC

|=|

OA

|=2，即△ABC為正三角形，
∴其面積S=

1

2

×2×2×sin60°=

3

點評：本題考查三角形形狀的判斷，涉及向量的數量積的運算，屬中檔題．