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
②在四面体ABCD中,若
③在四面体ABCD中点,且满足
④对空间任意点O与不共线的三点A,B,C,若
|
答案
①根据向量的加法法则可知
+AB
+BC
+CD
=DA
,∴①正确;0
②令
=AB
,a
=AD
,b
=AC
,c
∵
•AB
=0,CD
•AC
=0,BD
∴
⋅(AB
-AD
)=AC
⋅(a
-b
)=0,解得c
•a
=b
•a
①c
⋅(AC
-AD
)=AB
⋅(c
-b
)=0,解得a
⋅c
=b
⋅a
②,c
∴
⋅c
=b
⋅a
,即b
⋅c
-b
⋅a
=b
⋅(b
-c
)=0,a
即
⋅(AD
-AC
)=AB
⋅AD
=0,∴②正确.BC
③∵
=BC
-AC
,AB
=BD
-AD
,AB
•AB
=0,AC
•AC
=0,AD
•AB
=0.AD
∴
⋅BC
=(BD
-AC
)⋅(AB
-AD
)=AB
2>0,AB
∴∠CBD为锐角,同理
•CD
>0,CB
•DB
>0,DC
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
=x0P
+yOA
+zOB
(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.OC
故正确是个数有4个,
故选:D.