题目内容
已知O是△ABC内一点,且满足
•
=
•
=
•
,则O点一定是△ABC的( )
| OA |
| OB |
| OB |
| OC |
| OC |
| OA |
分析:利用
•
=
•
=
•
,由平面向量的运算法则,能推导出
⊥
,
⊥
,
⊥
,由此能求出结果.
| OA |
| OB |
| OB |
| OC |
| OC |
| OA |
| OB |
| CA |
| OA |
| CB |
| OC |
| AB |
解答:解:O是△ABC内一点,且满足
•
=
•
=
•
,
∴
•
-
•
=0,
∴
•(
-
)=
•
=0,
∴
⊥
,
∵
•
-
•
=0,
∴
•(
-
)=
•
=0,
∴
⊥
,
∵
•
-
•
=0,
∴
•(
-
)=
•
=0,
∴
⊥
,
∴O点一定是△ABC的垂心.
故选C.
| OA |
| OB |
| OB |
| OC |
| OC |
| OA |
∴
| OA |
| OB |
| OB |
| OC |
∴
| OB |
| OA |
| OC |
| OB |
| CA |
∴
| OB |
| CA |
∵
| OA |
| OB |
| OC |
| OA |
∴
| OA |
| OB |
| OC |
| OA |
| CB |
∴
| OA |
| CB |
∵
| OB |
| OC |
| OC |
| OA |
∴
| OC |
| OB |
| OA |
| OC |
| AB |
∴
| OC |
| AB |
∴O点一定是△ABC的垂心.
故选C.
点评:本题考查平面向量的应用,解题时要注意平面运算法则的灵活运用,是基础题.
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