平行六面体ABCD-A1B1C1D1中,AB=4,AD=3,AA1=5,角BAD=角BAA1=角DAA1=60°则AC1的长=?
问题描述:
平行六面体ABCD-A1B1C1D1中,AB=4,AD=3,AA1=5,角BAD=角BAA1=角DAA1=60°则AC1的长=?
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'*ABcos60+2AD*AA'cos60=16+9...