题目内容

如图,O是△ABC内一点,PQ∥BC,且
PQ
BC
=t,
OA
=
a
OB
=
b
OC
=
c
,试用
a
b
c
表示
OP
OQ
考点:平面向量的基本定理及其意义
专题:平面向量及应用
分析:
PQ
BC
=t,PQ∥BC,可得
AP
AB
=
AQ
AC
=t
,而
AP
=t
AB
AQ
=t
AC
.
AB
=
OB
-
OA
AC
=
OC
-
OA
.分别代入
OP
=
OA
+
AP
OQ
=
OA
+
AQ
,即可得出.
解答: 解:∵
PQ
BC
=t,PQ∥BC,∴
AP
AB
=
AQ
AC
=t

AP
=t
AB
AQ
=t
AC
.
AB
=
OB
-
OA
AC
=
OC
-
OA

OP
=
OA
+
AP
=
OA
+t(
OB
-
OA
)
=(1-t)
OA
+t
OB
=(1-t)
a
+t
b

OQ
=
OA
+
AQ
=
OA
+t(
OC
-
OA
)
=(1-t)
OA
+t
OC
=(1-t)
a
+t
c
点评:本题考查了向量的三角形法则、向量共线定理、平行线的性质,考查了推理能力与计算能力,属于中档题.
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