问题 填空题
已知|
OA
|=4,|
OB
|=2,∠AOB=
3
OC
=x
OA
+y
OB
,且x+2y=1,则|
OC
|
的最小值是
2
7
7
答案

∵x+2y=1

|

OC
|•|
OC
|
=(x
OA
+y
OB
)
2

=(1-2y)2×16+2y(1-2y)×2×4×(-

1
2
)+4y2

=84y2-72y+16

∴当y=

3
7
时,原式=
4
7

故答案为:

2
7
7

填空题
填空题