问题
填空题
已知|
|
答案
∵|
|=6,|a
|=6b
,(2
-a
)•(b
+3a
)=-108,b
∴36+2
•a
-216=-108b
∴
•a
=36b
∴cos<
,a
>=b
=36 6×6 2 2 2
∵<
,a
>∈[0,π]b
∴<
,a
>=b π 4
故答案为:π 4
已知|
|
∵|
|=6,|a
|=6b
,(2
-a
)•(b
+3a
)=-108,b
∴36+2
•a
-216=-108b
∴
•a
=36b
∴cos<
,a
>=b
=36 6×6 2 2 2
∵<
,a
>∈[0,π]b
∴<
,a
>=b π 4
故答案为:π 4