问题 解答题
已知向量|
a
|=|
b
|=1
,且
a
b
=-
1
2
,求:
(1)|
a
+
b
|

(2)
a
b
-
a
的夹角.
答案

(1)由题意可得|

a
+
b
|=
(
a
+
b
)2

=

a
2
+2
a
b
+
b
2
=
12+2×(-
1
2
)+12
=1;

(2)同理可得|

b
-
a
|=
12-2×(-
1
2
)+12
=
3

a
•(
b
-
a
)=
a
b
-
a
2
=-
1
2
-12
=-
3
2

故cos

a
b
-
a
>=
a
•(
b
-
a
)
|
a
||
b
-
a
|
=-
3
2

a
b
-
a
>∈[0,π],

a
b
-
a
的夹角
a
b
-
a
=
6

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