题目内容
| GB |
| GC |
(Ⅱ)已知向量
| a |
| 3x |
| 2 |
| 3x |
| 2 |
| b |
| x |
| 2 |
| x |
| 2 |
| a |
| b |
分析:(I)利用向量的运算法则和重心定理及数量积运算即可得出;
(II)利用模的计算公式和三角函数的平方关系及其两角和差的余弦公式即可得出.
(II)利用模的计算公式和三角函数的平方关系及其两角和差的余弦公式即可得出.
解答:解:(I)设
=
,
=
.
则
=
=
(
-
),
=
=
(
-
).
∴
•
=
(
-
)•
(
-
)
=
(
•
-
2-
2+
•
)
=
(
•
-
2-
2)
=
(
|
| |
|cos60°-
|
|2-
|
|2)
=
.
(II)∵|
+
|=
=
=
=1,
∴cos2x=-
,
又x∈[0,π],∴2x∈[0,2π].
∴2x=
或
,
解得x=
或x=
.
| AB |
| a |
| AC |
| b |
则
| GB |
| 2 |
| 3 |
| EB |
| 2 |
| 3 |
| a |
| 1 |
| 2 |
| b |
| GC |
| 2 |
| 3 |
| FC |
| 2 |
| 3 |
| b |
| 1 |
| 2 |
| a |
∴
| GB |
| GC |
| 2 |
| 3 |
| a |
| 1 |
| 2 |
| b |
| 2 |
| 3 |
| b |
| 1 |
| 2 |
| a |
=
| 4 |
| 9 |
| a |
| b |
| 1 |
| 2 |
| a |
| 1 |
| 2 |
| b |
| 1 |
| 4 |
| a |
| b |
=
| 4 |
| 9 |
| 5 |
| 4 |
| a |
| b |
| 1 |
| 2 |
| a |
| 1 |
| 2 |
| b |
=
| 4 |
| 9 |
| 5 |
| 4 |
| a |
| b |
| 1 |
| 2 |
| a |
| 1 |
| 2 |
| b |
=
| 5 |
| 9 |
(II)∵|
| a |
| b |
(cos
|
=
2+2(cos
|
=
| 2+2cos2x |
∴cos2x=-
| 1 |
| 2 |
又x∈[0,π],∴2x∈[0,2π].
∴2x=
| 2π |
| 3 |
| 4π |
| 3 |
解得x=
| π |
| 3 |
| 2π |
| 3 |
点评:熟练掌握向量的运算法则和重心定理及数量积运算、模的计算公式和三角函数的平方关系及其两角和差的余弦公式是解题的关键.
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