题目内容

在△ABC中,AC=2,BC=6,已知点O是△ABC内一点,且满足
OA
+3
OB
+4
OC
=
0
,则
OC
•(
BA
+2
BC
)
=
 
考点:平面向量数量积的运算
专题:平面向量及应用
分析:由于
OA
=
OC
+
CA
OB
=
OC
+
CB
,满足
OA
+3
OB
+4
OC
=
0
,可得
OC
=-
1
8
(
CA
+3
CB
)
.再利用数量积运算性质即可得出.
解答: 解:∵
OA
=
OC
+
CA
OB
=
OC
+
CB
,满足
OA
+3
OB
+4
OC
=
0

OC
+
CA
+3(
OC
+
CB
)
+4
OC
=
0

OC
=-
1
8
(
CA
+3
CB
)

BA
+2
BC
=
CA
-
CB
+2
BC
=
CA
-3
CB

OC
•(
BA
+2
BC
)
=-
1
8
(
CA
+3
CB
)•(
CA
-3
CB
)

=-
1
8
(
CA
2
-9
CB
2
)

=-
1
8
(22-62)

=40.
故答案为:40.
点评:本题考查了向量的运算法则、数量积运算性质,考查了推理能力与计算能力,属于难题.
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