平行六面体ABCD-A1B1C1D1中,AB=5AD＝3AA1＝√2∠BAD＝60∠BAA1＝∠DAA1E是CC1中点求AE长度

Question

平行六面体ABCD-A1B1C1D1中,AB=5 AD＝3 AA1＝√2 ∠BAD＝60 ∠BAA1＝∠DAA1 E是CC1中点 求AE长度

Answer 1

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'*ABcos60+2AD*AA'cos60
=
16+9+25+2*4*3/2+2*5*4/2+2*3*5/2
=
97
=>
AC' = 97