已知平行六面体ABCD－A′B′C′D′,AB＝5,AD＝3,AA′＝7,∠BAD＝60°,∠BAA′＝∠DAA′＝45°,求AC′的长．

Answer 1

均表示为向量
AC'=AB+ BC+ CC′=根号（AB+ BC+ CC′）^2
=根号[AB^2+BC^2+CC'^2+AB·BC+BC·CC'+AB·CC')
=根号[25+9+49+5*3*cos60+3*7*cos45+5*7*cos45]
=根号[83+15/2+28√2]
检验一下看算错没有,应该还能化简