已知平行六面体ABCD-A1B1C1D1中棱AB=3,AD=4,AA'=5,且角BAA1=DAA1=BAD=60°,求AC1的长?

Question

已知平行六面体ABCD-A1B1C1D1中棱AB=3,AD=4,AA'=5,且角BAA1=DAA1=BAD=60°,求AC1的长?

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 应该对的吧