题目内容
已知向量
=(3,4),
=(k,0)
(1)若
⊥(
-
),求k的值;
(2)若k=5,
与
-
所成的角为θ,求cosθ
| a |
| c |
(1)若
| a |
| a |
| c |
(2)若k=5,
| a |
| a |
| c |
分析:(1)由
=(3,4),
=(k,0),知
-
=(3-k,4),由
⊥(
-
),知3×(3-k)+4×4=0,由此能求出k.
(2)由k=5,知
-
=(-2,4),由此能求出
与
-
所成的角θ的余弦值.
| a |
| c |
| a |
| c |
| a |
| a |
| c |
(2)由k=5,知
| a |
| c |
| a |
| a |
| c |
解答:解:(1)∵
=(3,4),
=(k,0),
∴
-
=(3-k,4),(1分)
∵
⊥(
-
),∴
•(
-
)=0,(3分)
∴3×(3-k)+4×4=0,(5分)
解得k=
.(7分)
(2)∵k=5,∴
-
=(-2,4),(8分)
∴|
-
|=
=2
,
|
|=
=5,(11分)
cosθ=
=
=
.(14分)
| a |
| c |
∴
| a |
| c |
∵
| a |
| a |
| c |
| a |
| a |
| c |
∴3×(3-k)+4×4=0,(5分)
解得k=
| 25 |
| 3 |
(2)∵k=5,∴
| a |
| c |
∴|
| a |
| c |
| (-2)2+42 |
| 5 |
|
| a |
| 32+42 |
cosθ=
| ||||||
|
|
| -2×3+4×4 | ||
5×2
|
| ||
| 5 |
点评:本题考查平面向量的垂直关系的应用,考查平面向量的夹角的余弦值的求法,解题时要认真审题,仔细解答.
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