题目内容
| AB |
| a |
| AC |
| b |
| AF |
| a |
| b |
A、(
| ||||
B、(
| ||||
C、(
| ||||
D、(
|
分析:依题意可分别求得
和
,进而可分别表示出
和
,根据C,D,F共线和B,E,F共线分别表示出
,最后联立求得m和n,代入
=
+
求得x和y.
| AD |
| AC |
| CD |
| BE |
| AF |
| AF |
| AC |
| CF |
解答:解:依题意可知
=
,则
=
,
=
,则
=
,
=
-
=
-
,
=
-
=
-
,
∵C,D,F共线,
∴
=m
=
m
-m
=
+
=
m
+(1-m)
(1)
∵B,E,F共线,
∴
=n
=
n
-n
=
+
=
n
+(1-n)
(2)
又基底的分解形式唯一,由(1)(2)知
=1-n
1-m=
n
解得m=
,n=
代入(1),
=
+
=
m
+(1-m)
=
+
∴x=
,y=
故选C
| AB |
| a |
| AD |
| 2 |
| 3 |
| a |
| AC |
| b |
| AE |
| 3 |
| 4 |
| b |
| CD |
| AD |
| AC |
| 2 |
| 3 |
| a |
| b |
| BE |
| AE |
| AB |
| 3 |
| 4 |
| b |
| a |
∵C,D,F共线,
∴
| CF |
| CD |
| 2 |
| 3 |
| a |
| b |
| AF |
| AC |
| CF |
| 2 |
| 3 |
| a |
| b |
∵B,E,F共线,
∴
| BF |
| BE |
| 3 |
| 4 |
| b |
| a |
| AF |
| AB |
| BF |
| 3 |
| 4 |
| b |
| a |
又基底的分解形式唯一,由(1)(2)知
| 2m |
| 3 |
1-m=
| 3 |
| 4 |
解得m=
| 1 |
| 2 |
| 2 |
| 3 |
代入(1),
| AF |
| AC |
| CF |
| 2 |
| 3 |
| a |
| b |
| 1 |
| 3 |
| a |
| 1 |
| 2 |
| b |
∴x=
| 1 |
| 3 |
| 1 |
| 2 |
故选C
点评:本题主要考查了向量的三角形法则,向量的基本运算.考查了学生对向量基础知识的综合把握.
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