题目内容

设O、A、B、C为平面内四点,
OA
=
a
OB
=
b
OC
=
c
,且
a
+
b
+
c
=
0
a
b
=
b
c
=
c
a
=-1
,则|
a
|2+|
b
|2+|
c
|2
=______.
a
+
b
+
c
=
0
,平方得:
|
a
|2+|
b
|2+|
c
|2+2(
a
b
+
b
c
+
c
a
)=0

a
b
=
b
c
=
c
a
=-1

|
a
|2+|
b
|2+|
c
|2+2×(-3)=0

|
a
|2+|
b
|2+|
c
|2=6

故答案为:6.
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