如图,平行六面体ABCD-A'B'C'D'中,AB=5,AD=3,AA'=7,∠BAD=60°∠BAA'=∠DAA'=45°,求AC'的长.
问题描述：
如图,平行六面体ABCD-A'B'C'D'中,AB=5,AD=3,AA'=7,∠BAD=60°∠BAA'=∠DAA'=45°,求AC'的长.
最好具体点.Thanks...
答
向量AC'=向量AB+向量AD+向量AA'
=>
AC'^2 = (向量AB+向量AD+向量AA')^2
=
AB^2 + AD^2 + AA'^2 + 2(向量AB*向量AD+向量AA'*向量AB+向量AD*向量AA')
=
AB^2 + AD^2 + AA'^2 + 2AB*ADcos60+2AA'*ABcos45+2AD*AA'cos45
=
25+9+49+15+35√2+21√2
=
98+56√2
=>
AC' = √(98+56√2)