题目内容

设O、A、B、C为平面上四个点,
OA
=
a
OB
=
b
OC
=
c
,且
a
+
b
+
c
=
0
a
b
=
b
c
=
c
a
=-1
,则|
a
|+|
b
|+|
c
|
等于(  )
A.2
2
B.2
3
C.3
2
D.3
3
a
+
b
+
c
=
0
a
b
=
b
c
=
c
a
=-1
,∴
a
2
+
b
2
+
c
2
-6=0,
a
+
b
=-
c
 两边平方得 a2+
b
2
-2=
c
2
,∴
c
2
=2,∴|
c
|=
2
,a2+
b
2
=4,
a
+
c
=-
b
 两边平方得 a2+
c
2
-2=
b
2
,∴a2+2-2=
b
2
,∴a2=
b
2
=2,
∴|
a
|=|
b
|=
2
,则|
a
|+|
b
|+|
c
|
=3
2

故选 C.
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