问题 选择题
下列命题中真命题的个数是(  )
①若A,B,C,D是空间任意四点,则有
AB
+
BC
+
CD
+
DA
=
0

②在四面体ABCD中,若
AB
CD
=0,
AC
BD
=0
,则
AD
BC
=0

③在四面体ABCD中点,且满足
AB
AC
=0,
AC
AD
=0
AB
AD
=0
.则△BDC是锐角三角形
④对空间任意点O与不共线的三点A,B,C,若
OP
=x
OA
+y
OA
+z
OC
(其中x,y,z∈R且x+y+z=1),则P,A,B,C四点共面.
A.1B.2C.3D.4
答案

①根据向量的加法法则可知

AB
+
BC
+
CD
+
DA
=
0
,∴①正确;

②令

AB
=
a
AD
=
b
AC
=
c

AB
CD
=0,
AC
BD
=0,

AB
⋅(
AD
-
AC
)=
a
⋅(
b
-
c
)=0,解得
a
b
=
a
c

AC
⋅(
AD
-
AB
)=
c
⋅(
b
-
a
)=0,解得
c
b
=
a
c
②,

c
b
=
a
b
,即
c
b
-
a
b
=
b
⋅(
c
-
a
)=0

AD
⋅(
AC
-
AB
)=
AD
BC
=0,∴②正确.

③∵

BC
=
AC
-
AB
BD
=
AD
-
AB
AB
AC
=0,
AC
AD
=0
AB
AD
=0

BC
BD
=(
AC
-
AB
)⋅(
AD
-
AB
)=
AB
2
>0

∴∠CBD为锐角,同理

CD
CB
>0,
DB
DC
>0,

即△BDC是锐角三角形,∴③正确.

④对空间任意点O与不共线的三点A、B、C,若

0P
=x
OA
+y
OB
+z
OC
(其中x、y、z∈R),只有当x+y+z=1时,P、A、B、C四点才共面,∴④正确.

故正确是个数有4个,

故选:D.

单项选择题
单项选择题