题目内容
如图,已知矩形ABCD中,|AD|=3,|AB|=4.将矩形ABCD沿对角线BD折起,使得面BCD⊥面ABD.现以D为原点,DB作为y轴的正方向,建立如图空间直角坐标系,此时点A恰好在xDy坐标平面内.试求A,C两点的坐标.

解:如图,由于面BCD⊥面ABD,从面BCD引棱DB的垂线CF即为面ABD的垂线,同理可得AE即为面BCD的垂线
∵矩形ABCD中,|AD|=3,|AB|=4,∴BD=5
在直角三角形DAB与直角三角形DCB中,由射影定理知
DA2=DE×BD,即9=DE×5,得DE=
BC2=BF×BD,即9=BF×5得BF=
由勾股定理可解得CF=AE=
,故EF=5-DE-BF=5-
-
=
∴DF=DE+EF=
+
=
故在空间坐标系中,得A,C两点的坐标为A(
),C(0,
)
分析:由于面BCD⊥面ABD,从面BCD引棱DB的垂线CF即为面ABD的垂线,同理可得AE即为面BCD的垂线,故只需求得AE,CF,DE,DF的长度即可.
点评:本题考点是空间坐标系,考查求空间坐标系中点的坐标的方法,及坐标符号正负的确定.
∵矩形ABCD中,|AD|=3,|AB|=4,∴BD=5
在直角三角形DAB与直角三角形DCB中,由射影定理知
DA2=DE×BD,即9=DE×5,得DE=

BC2=BF×BD,即9=BF×5得BF=

由勾股定理可解得CF=AE=




∴DF=DE+EF=



故在空间坐标系中,得A,C两点的坐标为A(


分析:由于面BCD⊥面ABD,从面BCD引棱DB的垂线CF即为面ABD的垂线,同理可得AE即为面BCD的垂线,故只需求得AE,CF,DE,DF的长度即可.
点评:本题考点是空间坐标系,考查求空间坐标系中点的坐标的方法,及坐标符号正负的确定.

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