题目内容
已知ABCDEF是正六边形,且
=
,
=
,则
=( )
| AB |
| a |
| AE |
| b |
| CD |
分析:正六边形ABCDEF中,根据
=
=
(
-
),且
=
,
=
,由此得到结论.
| CD |
| 1 |
| 2 |
| BE |
| 1 |
| 2 |
| AE |
| AB |
| AB |
| a |
| AE |
| b |
解答:
解:如图,在正六边形ABCDEF中,由正六边形的性质可得
=
=
(
-
)=
(
-
),
故选B.
| CD |
| 1 |
| 2 |
| BE |
| 1 |
| 2 |
| AE |
| AB |
| 1 |
| 2 |
| b |
| a |
故选B.
点评:本题考查向量相等的概念、向量加减法的三角形及平行四边形法则、向量共线的充要条件,体现了数形结合的数学思想,属于基础题题.
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