题目内容
已知向量
,
是夹角为60°的两个单位向量,点C,D满足
=
=
,动点P满足
•
=0,且
=x
+y
(x,y∈R),则xy的最大值为 .
| OA |
| OB |
| AC |
. |
| CD |
| DB |
| DP |
| OC |
| OP |
| OA |
| OB |
考点:平面向量数量积的运算
专题:计算题,平面向量及应用
分析:由
=
=
,知C、D为线段AB的三等分点,从而可用向量
,
表示向量
,
,则
•
=0,可化为(
-
)•
=0,即(x
+y
-
-
)•(
+
)=0,利用向量数量积运算整理可得5x+4y-
=0,消掉y可把xy化为x的二次函数,由二次函数性质可求答案.
| AC |
. |
| CD |
| DB |
| OA |
| OB |
| OC |
| OD |
| DP |
| OC |
| OP |
| OD |
| OC |
| OA |
| OB |
| 1 |
| 3 |
| OA |
| 2 |
| 3 |
| OB |
| 2 |
| 3 |
| OA |
| 1 |
| 3 |
| OB |
| 13 |
| 3 |
解答:
解:∵
,
是夹角为60°的两个单位向量,
∴
•
=
,
∵
=
=
,
∴C、D为线段AB的三等分点,
=
+
=
+
(
-
)=
+
,
=
+
=
+
(
-
)=
+
,
则
•
=0,即(
-
)•
=0,
∴(x
+y
-
-
)•(
+
)=0,
∴5x+4y-
=0,
则xy=x(
-
x)=-
(x-
)2+
,
∴当x=
时xy取得最大值
,
故答案为:
.
| OA |
| OB |
∴
| OA |
| OB |
| 1 |
| 2 |
∵
| AC |
. |
| CD |
| DB |
∴C、D为线段AB的三等分点,
| OC |
| OA |
| 1 |
| 3 |
| AB |
| OA |
| 1 |
| 3 |
| OB |
| OA |
| 2 |
| 3 |
| OA |
| 1 |
| 3 |
| OB |
| OD |
| OA |
| 2 |
| 3 |
| AB |
| OA |
| 2 |
| 3 |
| OB |
| OA |
| 1 |
| 3 |
| OA |
| 2 |
| 3 |
| OB |
则
| DP |
| OC |
| OP |
| OD |
| OC |
∴(x
| OA |
| OB |
| 1 |
| 3 |
| OA |
| 2 |
| 3 |
| OB |
| 2 |
| 3 |
| OA |
| 1 |
| 3 |
| OB |
∴5x+4y-
| 13 |
| 3 |
则xy=x(
| 13 |
| 12 |
| 5 |
| 4 |
| 5 |
| 4 |
| 13 |
| 30 |
| 169 |
| 720 |
∴当x=
| 13 |
| 30 |
| 169 |
| 720 |
故答案为:
| 169 |
| 720 |
点评:本题考查平面向量基本定理、向量数量积运算及二次函数的性质,考查学生灵活运用知识解决问题的能力.
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