题目内容
| AB |
| a |
| AC |
| b |
(1)试用
| a |
| b |
| AD |
(2)求
| AD |
| BC |
分析:(1)根据题意得
=
,由向量的减法法则得
=
-
,从而可得
=
+
=
+
=
+
;
(2)由(1)可得:
•
=(
+
)•(
-
)=
2+
•
-
2,根据题意算出
•
=-1,
2=4且
2=1,代入加以计算即可得到
•
的值.
| BD |
| 1 |
| 3 |
| BC |
| BC |
| b |
| a |
| AD |
| AB |
| BD |
| AB |
| 1 |
| 3 |
| BC |
| 2 |
| 3 |
| a |
| 1 |
| 3 |
| b |
(2)由(1)可得:
| AD |
| BC |
| 2 |
| 3 |
| a |
| 1 |
| 3 |
| b |
| b |
| a |
| 1 |
| 3 |
| b |
| 1 |
| 3 |
| a |
| b |
| 2 |
| 3 |
| a |
| a |
| b |
| a |
| b |
| AD |
| BC |
解答:解:(1)∵D是边BC上一点,DC=2BD,∴
=
,
又∵
=
,
=
,
=
-
,
∴
=
+
=
+
=
+
(
-
)=
+
.
(2)∵|
|=|
|=2,|
|=|
|=1,∠BAC=120°,
∴
•
=|
|•|
|cos∠BAC=2×1×120°=-1,
因此,
•
=(
+
)•(
-
)
=
2+
•
-
2=
×12+
×(-1)-
×22=-
.
| BD |
| 1 |
| 3 |
| BC |
又∵
| AB |
| a |
| AC |
| b |
| BC |
| b |
| a |
∴
| AD |
| AB |
| BD |
| AB |
| 1 |
| 3 |
| BC |
| a |
| 1 |
| 3 |
| b |
| a |
| 2 |
| 3 |
| a |
| 1 |
| 3 |
| b |
(2)∵|
| a |
| AB |
| b |
| AC |
∴
| a |
| b |
| a |
| b |
因此,
| AD |
| BC |
| 2 |
| 3 |
| a |
| 1 |
| 3 |
| b |
| b |
| a |
=
| 1 |
| 3 |
| b |
| 1 |
| 3 |
| a |
| b |
| 2 |
| 3 |
| a |
| 1 |
| 3 |
| 1 |
| 3 |
| 2 |
| 3 |
| 8 |
| 3 |
点评:本题在特殊的三角形中求向量的数量积.着重考查了平面向量的线性运算法则、向量的数量积及其运算性质等知识,属于中档题.
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