题目内容
已知空间任一点O和不共线的三点A,B,C,满足
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A.充分但不必要条件
B.必要但不充分条件
C.充要条件
D.既不充分也不必要条件
【答案】分析:要寻求四点A、B、C、D共面的充要条件,自然想到共面向量定理.用
表示出
,进而用
表示
,三者的系数之和为1即可找出答案.
解答:解:已知空间任一点O和不共线的三点A,B,C,满足
是“点P位于平面ABC内”的充要条件.证明如下:
(必要性)依题意知,B、C、D三点不共线,
则由共面向量定理的推论知:四点A、B、C、D共面
?对空间任一点O,存在实数x1、y1,使得
=
+x1
+y1 
=
+x1(
-
)+y1(
-
)
=(1-x1-y1)
+x1
+y1
,
取x=1-x1-y1、y=x1、z=y1,
则有
=x
+y
+z
,且x+y+z=1.
(充分性)对于空间任一点O,存在实数x、y、z且x+y+z=1,使得
=x
+y
+z
.
所以x=1-y-z得
=(1-y-z)
+y
+z
.
=
+y
+z
,即:
,
所以四点A、B、C、D共面.
所以,空间任意无三点共线的四点A、B、C、D共面的充分必要条件是:
对于空间任一点O,存在实数x、y、z且x+y+z=1,使得
=x
+y
+z
.
故选C.
点评:本题考查共线向量与共面向量定理,考查学生分析问题解决问题的能力,是中档题.
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解答:解:已知空间任一点O和不共线的三点A,B,C,满足
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(必要性)依题意知,B、C、D三点不共线,
则由共面向量定理的推论知:四点A、B、C、D共面
?对空间任一点O,存在实数x1、y1,使得
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=
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=(1-x1-y1)
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取x=1-x1-y1、y=x1、z=y1,
则有
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(充分性)对于空间任一点O,存在实数x、y、z且x+y+z=1,使得
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所以x=1-y-z得
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所以四点A、B、C、D共面.
所以,空间任意无三点共线的四点A、B、C、D共面的充分必要条件是:
对于空间任一点O,存在实数x、y、z且x+y+z=1,使得


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
故选C.
点评:本题考查共线向量与共面向量定理,考查学生分析问题解决问题的能力,是中档题.
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已知空间任一点O和不共线的三点A,B,C,满足
=x
+y
+z
(x,y,z∈R),则”x+y+z=1”是“点P位于平面ABC内”的( )
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