问题 填空题
平面上的向量
PA
PB
满足
PA
2
+
PB
2
=4,且
PA
PB
=0
,若向量
PC
=
1
3
PA
+
2
3
PB
,则|
PC
|

最大为______.
答案

向量

PA
PB
满足
PA
2
+
PB
2
=4,且
PA
PB
=0

∵向量

PC
 = 
1
3
PA
+
2
3
PB

|

PA
|=x  , |
PB
|=y,则x2+y2=4

|

PC
|=
(
1
3
PA
 +
2
3
PB
) 2
=
1
9
PA
2
+
4
9
 
PB
2

=

1
9
x
2
+
4
9
y2
=
x2
9
+
4(4-x2)
9

=

-
1
3
x2+
16
9
 

当x=0时 |

PC
|=
4
3
为最大值

故答案为:

4
3

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