题目内容
已知△OAB中,点D在线段OB上,且OD=2DB,延长BA到C,使BA=AC.设
=
,
=
.
(1)用
,
表示向量
,
;
(2)若向量
与
+k
共线,求k的值.
| OA |
| a |
| OB |
| b |
(1)用
| a |
| b |
| OC |
| DC |
(2)若向量
| OC |
| OA |
| DC |
(1)∵A为BC的中点,∴
=
(
+
),
可得
=2
-
=2
-
,
而
=
-
=
-
=2
-
(2)由(1),得
+k
=(2k+1)
-
k
,
∵
与
+k
共线,设
=λ(
+k
)
即2
-
=λ(2k+1)
+-
λk
,
根据平面向量基本定理,得
解之得,k=
.
| OA |
| 1 |
| 2 |
| OB |
| OC |
可得
| OC |
| OA |
| OB |
| a |
| b |
而
| DC |
| OC |
| OD |
| OC |
| 2 |
| 3 |
| OB |
| a |
| 5 |
| 3 |
| b |
(2)由(1),得
| OA |
| DC |
| a |
| 5 |
| 3 |
| b |
∵
| OC |
| OA |
| DC |
| OC |
| OA |
| DC |
即2
| a |
| b |
| a |
| 5 |
| 3 |
| b |
根据平面向量基本定理,得
|
解之得,k=
| 3 |
| 4 |
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