四棱柱ABCD-A1B1C1D1的底面ABCD为矩形,AB=1,AD=2,AA1=3,角A1AB=角A1AD=60度,求AC1._数学解答

四棱柱ABCD-A1B1C1D1的底面ABCD为矩形,AB=1,AD=2,AA1=3,角A1AB=角A1AD=60度,求AC1.

人气：181 ℃　时间：2019-12-01 08:43:22

解答

AC1=AB+AD+AA1
∴ |AC1|²
=（AB+AD+AA1）²
=AB²+AD²+AA1²+2AB.AD +2AB.AA1+2AD.AA1
=1+4+9+0+2*1*3*(1/2)+2*2*3*(1/2)
=14+3+6
=23
∴ |AC|=√23

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