取A为原点,AB为x轴.AC为y轴,AD为z轴,
平面BCD方程:x/1+y/2+z/4=1.
法线式为:(4x+2y+z-4)/√21=0.
A到平面BCD距离=|4×0+2×0+0-4|/√21=4/√21≈0.873.
在四面体ABCD中,AB,AC,AD两两垂直,且AB=1,AC=2,AD=4,求点A到平面BCD距离
在四面体ABCD中,AB,AC,AD两两垂直,且AB=1,AC=2,AD=4,求点A到平面BCD距离
数学人气:701 ℃时间:2019-10-19 14:42:31
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