题目内容
在△ABC中,D是BC边上一点,BD=3DC,若P是AD边上一动点,AD=2
(Ⅰ)设
=
,
=
,用
,
表示向量
(Ⅱ)求
•(
+3
)的最小值.
(Ⅰ)设
| PB |
| a |
| PC |
| b |
| a |
| b |
| PD |
(Ⅱ)求
| PA |
| PB |
| PC |
(Ⅰ)依题
=
-
,
=
-
又
=-3
所以
-
=-3(
-
)
整理可得4
=
+3
则
=
+
(Ⅱ)由(Ⅰ)可知
+3
=4
设|
|=x(0≤x≤2)故
•(
+3
)=
•(4
)=-4x(2-x)≥-4
所以当x=1时
•(
+3
)的最小值为-4
| BD |
| PD |
| PB |
| CD |
| PD |
| PC |
又
| BD |
| CD |
| PD |
| PB |
| PD |
| PC |
整理可得4
| PD |
| PB |
| PC |
| PD |
| 1 |
| 4 |
| a |
| 3 |
| 4 |
| b |
(Ⅱ)由(Ⅰ)可知
| PB |
| PC |
| PD |
设|
| PA |
| PA |
| PB |
| PC |
| PA |
| PD |
所以当x=1时
| PA |
| PB |
| PC |
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