高二数学题,必修2的在长方体ABCD—A1B1C1D1中AA1＝a,∠BAB1＝∠B1A1C1＝30°,则异面直线AB1

AA1 = a => BB1 = a => AB1 = a / sin(30) = 2a
=> A1B1 = AB = a / tan(30) = sq(3)a (sq(3)代表根号3)
=> AC = A1C1 = A1B1 / cos(30) = 2a,B1C1 = A1B1 × tan(30) = a
=> B1C = sq(B1B^2 + B1C1^2) = sq(2)a
异面直线AB1与A1C1的夹角为∠B1AC（因为A1C1与AC平行）
根据余弦定理,得
cos(∠B1AC) = (AB1^2 + AC^2 - B1C^2) / (2×AB1×AC) ＝ 2 / 3
注：AB1^2 代表AB1的平方