已知O,A,B是平面上的三个点,直线AB上有一点C,满足2AC+CB=0,则OC等于(  ) A.2OA-OB B.-OA+2OB C.23OA-13OB D.-13OA+23OB

问题描述:

已知O,A,B是平面上的三个点,直线AB上有一点C,满足2

AC
+
CB
=0,则
OC
等于(  )
A. 2
OA
-
OB

B. -
OA
+2
OB

C.
2
3
OA
-
1
3
OB

D. -
1
3
OA
+
2
3
OB

∵依题

OC
=
OB
+
BC
=
OB
+2
AC
=
OB
+2(
OC
-
OA
).
OC
=2
OA
-
OB
.

故选A