已知a=(1,1,0) b=(1,0,1)已知单位向量c与a,b共面,且c垂直b,求向量c
1个回答
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解:向量a
=
(1,1,0),向量b
=
(1,0,1),设单位向量c
=
(x,y,z),则√(x
2
+
y
2
+
z
2
)
=
1
=>
(x
2
+
y
2
+
z
2
)
=
1①,因为向量c⊥b,所以c·b
=
0
=>
(x,y,z)·(1,0,1)
=
x
+
z
=
0②,而向量c与a,b共面,说明存在k,t使得c
=
ka
+
tb
=
k(1,1,0)
+
t(1,0,1)
=
(k
+
t,k,t),这样有x
=
k
+
t,y
=
k,z
=
t
=>
x
=
y
+
z③,联立解得x
=
-z,y
=
-2z代入①,(-z)
2
+
(-2z)
2
+
z
2
=
1
=>
6z
2
=
1
=>
z
2
=
1/6
=>
z
=±√6/6,所以向量c
=
(-
√6/6
,
-
√6/3
,
√6/6)
或者
(
√6/6
,
√6/3
,
-
√6/6)
。
=
(1,1,0),向量b
=
(1,0,1),设单位向量c
=
(x,y,z),则√(x
2
+
y
2
+
z
2
)
=
1
=>
(x
2
+
y
2
+
z
2
)
=
1①,因为向量c⊥b,所以c·b
=
0
=>
(x,y,z)·(1,0,1)
=
x
+
z
=
0②,而向量c与a,b共面,说明存在k,t使得c
=
ka
+
tb
=
k(1,1,0)
+
t(1,0,1)
=
(k
+
t,k,t),这样有x
=
k
+
t,y
=
k,z
=
t
=>
x
=
y
+
z③,联立解得x
=
-z,y
=
-2z代入①,(-z)
2
+
(-2z)
2
+
z
2
=
1
=>
6z
2
=
1
=>
z
2
=
1/6
=>
z
=±√6/6,所以向量c
=
(-
√6/6
,
-
√6/3
,
√6/6)
或者
(
√6/6
,
√6/3
,
-
√6/6)
。
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