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A,B,O是平面内不共线的三个定点,OA=a,OB=b,点P关于点A的对称点为Q,点Q关于点B的对称点为R,则PR等于()A.a-bB.2(b-a)C.2(a-b)D.b-a

题目详情

A,B,O是平面内不共线的三个定点,

OA
=
a
OB
=
b
,点P关于点A的对称点为Q,点Q关于点B的对称点为R,则
PR
等于(  )

A.

a
-
b

B. 2(

b
-
a

C. 2(

a
-
b

D.

b
-
a

A,B,O是平面内不共线的三个定点,

OA
=
a
OB
=
b
,点P关于点A的对称点为Q,点Q关于点B的对称点为R,则
PR
等于(  )

A,B,O是平面内不共线的三个定点,
OA
=
a
OB
=
b
,点P关于点A的对称点为Q,点Q关于点B的对称点为R,则
PR
等于(  )
OA
=
a
OB
=
b
,点P关于点A的对称点为Q,点Q关于点B的对称点为R,则
PR
等于(  )
OA
OA
OAOA
a
OB
=
b
,点P关于点A的对称点为Q,点Q关于点B的对称点为R,则
PR
等于(  )
a
a
aa
OB
=
b
,点P关于点A的对称点为Q,点Q关于点B的对称点为R,则
PR
等于(  )
OB
OB
OBOB
b
,点P关于点A的对称点为Q,点Q关于点B的对称点为R,则
PR
等于(  )
b
b
bb
PR
等于(  )
PR
PR
PRPR

A.

a
-
b

a
-
b
a
a
aa
b
b
b
bb

B. 2(

b
-
a

b
-
a
b
b
bb
a
a
a
aa

C. 2(

a
-
b

a
-
b
a
a
aa
b
b
b
bb

D.

b
-
a

b
-
a
b
b
bb
a
a
a
aa
▼优质解答
答案和解析
根据向量的平行四边形法则得
2
a
=
OP
+
OQ
,2
b
=
OQ
+
OR

∴2(
b
-
a
)=
OR
-
OP
=
PR

故选:B.
a
a
aaa=
OP
+
OQ
,2
b
=
OQ
+
OR

∴2(
b
-
a
)=
OR
-
OP
=
PR

故选:B.
OP
OP
OPOPOP+
OQ
,2
b
=
OQ
+
OR

∴2(
b
-
a
)=
OR
-
OP
=
PR

故选:B.
OQ
OQ
OQOQOQ,2
b
=
OQ
+
OR

∴2(
b
-
a
)=
OR
-
OP
=
PR

故选:B.
b
b
bbb=
OQ
+
OR

∴2(
b
-
a
)=
OR
-
OP
=
PR

故选:B.
OQ
OQ
OQOQOQ+
OR

∴2(
b
-
a
)=
OR
-
OP
=
PR

故选:B.
OR
OR
OROROR,
∴2(
b
-
a
)=
OR
-
OP
=
PR

故选:B.
b
b
bbb-
a
)=
OR
-
OP
=
PR

故选:B.
a
a
aaa)=
OR
-
OP
=
PR

故选:B.
OR
OR
OROROR-
OP
=
PR

故选:B.
OP
OP
OPOPOP=
PR

故选:B.
PR
PR
PRPRPR
故选:B.