题目内容

如图,在△OAB中,已知P为线段AB上的一点,
OP
=x•
OA
+y•
OB

(1)若
BP
=
PA
,求x,y的值;
(2)若
BP
=3
PA
|
OA
|=4
|
OB
|=2
,且
OA
OB
的夹角为60°时,求
OP
AB
的值.
精英家教网
(1)∵
BP
=
PA

BO
+
OP
=
PO
+
OA
,即2
OP
=
OB
+
OA

OP
=
1
2
OA
+
1
2
OB
,即x=
1
2
y=
1
2

(2)∵
BP
=3
PA

BO
+
OP
=3
PO
+3
OA
,即4
OP
=
OB
+3
OA

OP
=
3
4
OA
+
1
4
OB

x=
3
4
y=
1
4

OP
AB
=(
3
4
OA
+
1
4
OB
)•(
OB
-
OA
)

=
1
4
OB
OB
-
3
4
OA
OA
+
1
2
OA
OB

=
1
4
×22-
3
4
×42+
1
2
×4×2×
1
2
=-9
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网