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.
问题描述：
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.
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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