问题 解答题
已知|
a
|=4
|
b
|=3
(2
a
-3
b
)(2
a
+
b
)=61

(1)求
a
b
的夹角θ;
(2)求|
a
+
b
|

(3)若
AB
=
a
BC
=
b
,求△ABC的面积.
答案

(1)∵(2

a
-3
b
)(2
a
+
b
)=61,∴4|
a
|2-4
a
b
-3|
b
|2=61,

又|

a
|=4,|
b
|=3,∴64-4
a
b
-27=61,∴
a
b
=-6,

cosθ=

a•b
|a||b|
=
-6
4×3
=-
1
2

又0≤θ≤π,

θ=

3

(2)|a+b|=

(a+b)2
=
|a|2+2a•b+|b|2
=
13

(3)∵

AB
BC
的夹角θ=
3

∠ABC=π-

3
=
π
3

|

AB
|=|a|=4,|
BC
|=|b|=3

S△ABC=

1
2
|
AB
||
BC
|sin∠ABC=
1
2
×4×3×
3
2
=3
3

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