在正方体ABCD-A1B1C1D1中,M是AB的中点,求异面直线A1C1与B1M所成角的余弦值

取CD的中点N,连接NC1.容易证明C1N//B1M.
即角A1C1N=所求角.
在三角形A1C1N中:A1C1=(根号2)*a,
C1N=(根号5)/2
A1N=根号[a^2+a^2+(0.5a)^2]
=1.5*a.
由余弦定理:
cos(角A1C1N)=
=[2+5/4-2.25]/[2*(根号2)*(根号5)/2]
=1/根号10.