在平行六面体ABCD-A'B'C'D'中,AB=4,AD=3,AA'=5,角BAD=90度,角BAA'=角DAA'=60度,求AC'的长
问题描述：
在平行六面体ABCD-A'B'C'D'中,AB=4,AD=3,AA'=5,角BAD=90度,角BAA'=角DAA'=60度,求AC'的长
答
用向量法
向量AC'=AC+CC'=AB+BC+CC',
AC'^2=AB^2+BC^2+CC'^2+2AB·BC+2AB·CC’+2BC·CC’
=|AB|^2+|BC|^2+|CC'|^2+2|AB||BC*cos90°+2|AB||CC'|cos60°+2|BC||CC'|cos60°
=16+9+25+2*4*3*0+2*4*5*1/2+2*3*5*1/2
=85,
|AC'|=√85