题目内容
已知向量|
|=1,|
|=2,|
|=3,且
,
,
两两的夹角都是
π,求:
(1)(2
+3
)•(
+2
);
(2)|
+
+
|;
(3)
+
+
与
所成的夹角.
| a |
| b |
| c |
| a |
| b |
| c |
| 2 |
| 3 |
(1)(2
| a |
| c |
| b |
| c |
(2)|
| a |
| b |
| c |
(3)
| a |
| b |
| c |
| c |
考点:数量积表示两个向量的夹角
专题:平面向量及应用
分析:(1)由题意可得
•
=-1,
•
=-3,
•
=-
.再根据(2
+3
)•(
+2
)=2
•
+4
•
+3
•
+6c2,计算求得结果.
(2)|
+
+
|=
=
,计算求得结果.
(3)设
+
+
与
所成的夹角为θ,则由cosθ=
,求得θ 的值.
| a |
| b |
| b |
| c |
| a |
| c |
| 3 |
| 2 |
| a |
| c |
| b |
| c |
| a |
| b |
| a |
| c |
| b |
| c |
(2)|
| a |
| b |
| c |
(
|
|
(3)设
| a |
| b |
| c |
| c |
(
| ||||||||
|
|
解答:
解:(1)由题意可得
•
=1×2×cos
=-1,
•
=2×3×cos
=-3,
•
=1×3×cos
=-
.
∴(2
+3
)•(
+2
)=2
•
+4
•
+3
•
+6c2=-2-6-9+54=37.
(2)|
+
+
|=
=
=
=
.
(3)设
+
+
与
所成的夹角为θ,则cosθ=
=
=
,
∴θ=
.
| a |
| b |
| 2π |
| 3 |
| b |
| c |
| 2π |
| 3 |
| a |
| c |
| 2π |
| 3 |
| 3 |
| 2 |
∴(2
| a |
| c |
| b |
| c |
| a |
| b |
| a |
| c |
| b |
| c |
(2)|
| a |
| b |
| c |
(
|
|
=
| 1+4+9-2-6-3 |
| 3 |
(3)设
| a |
| b |
| c |
| c |
(
| ||||||||
|
|
-
| ||
|
| ||
| 2 |
∴θ=
| π |
| 6 |
点评:本题主要考查两个向量的数量积的定义,数量积表示两个两个向量的夹角,属于中档题.
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