题目内容
在正方体ABCD—A1B1C1D1中,若E为A1C1中点,则直线CE垂直于( )
A.AC | B.BD | C.A1D | D.A1A |
B
分析:建立空间直角坐标系,设正方体棱长为1,求出向量
的坐标,以及
、
、
的坐标,可以发现
?
=0,因此,
⊥
,即CE⊥BD,
解答:解:以A为原点,AB、AD、AA1所在直线分别为x,y,z轴建空间直角坐标系,设正方体棱长为1,
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则A(0,0,0),C(1,1,0),B(1,0,0),
D(0,1,0),A1(0,0,1),E(
,
,1),
∴
=(-
,-
,1),
=(1,1,0),
=(-1,1,0),
=(0,1,-1),
=(0,0,-1),
显然
?
=
-
+0=0,
∴
⊥
,即CE⊥BD.
故选 B.
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解答:解:以A为原点,AB、AD、AA1所在直线分别为x,y,z轴建空间直角坐标系,设正方体棱长为1,

则A(0,0,0),C(1,1,0),B(1,0,0),
D(0,1,0),A1(0,0,1),E(
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∴
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显然
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∴
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故选 B.
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