题目内容
已知:
=(x,4,1),
=(-2,y,-1),
=(3,-2,z),
∥
,
⊥
,求:
(1)
,
,
;
(2)(
+
)与(
+
)所成角的余弦值.
a |
b |
c |
a |
b |
b |
c |
(1)
a |
b |
c |
(2)(
a |
c |
b |
c |
(1)∵
∥
,∴
=
=
,解得x=2,y=-4,
故
=(2,4,1),
=(-2,-4,-1),
又因为
⊥
,所以
•
=0,即-6+8-z=0,解得z=2,
故
=(3,-2,2)
(2)由(1)可得
+
=(5,2,3),
+
=(1,-6,1),
设向量
+
与
+
所成的角为θ,
则cosθ=
=-
a |
b |
x |
2 |
4 |
y |
1 |
-1 |
故
a |
b |
又因为
b |
c |
b |
c |
故
c |
(2)由(1)可得
a |
c |
b |
c |
设向量
a |
c |
b |
c |
则cosθ=
5-12+3 | ||||
|
2 |
19 |
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