题目内容
已知O为△ABC的重心,设
=
,
=
,则
=( )
| AB |
| a |
| AC |
| b |
| OB |
分析:首先由O是△ABC的重心,可得BO=
BE,利用向量加法的形法则,即可求得
=
=
(
-
)=
(
-
),化简可得结果.
| 2 |
| 3 |
| OB |
| 2 |
| 3 |
| EB |
| 2 |
| 3 |
| AB |
| AE |
| 2 |
| 3 |
| AB |
| 1 |
| 2 |
| AC |
解答:解:如图所示:设△ABC的中线AD 和BE相交于点O,由O为△ABC的重心可得,BO=
BE,
∴
=
=
(
-
)=
(
-
)=
(
-
)=
-
,
故选C.

| 2 |
| 3 |
∴
| OB |
| 2 |
| 3 |
| EB |
| 2 |
| 3 |
| AB |
| AE |
| 2 |
| 3 |
| AB |
| 1 |
| 2 |
| AC |
| 2 |
| 3 |
| a |
| 1 |
| 2 |
| b |
| 2 |
| 3 |
| a |
| 1 |
| 3 |
| b |
故选C.
点评:此题考查了两个向量的加减法的法则,以及其几何意义,三角形重心的性质,解此题的关键是数形结合思想的应用,属于基础题.
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