Question

正方体ABCD-A1B1C1D1中,点M为棱CC1的中点,求二面角A1-BD-M的余弦值

Answer 1

解：连接AC交BD于点O,连接A1O,OM,A1M

∵A1B=A1O,BM=DM,且点O为BD的中点，∴A1O⊥BD,OM⊥BD，∴∠A1OM即为所求角

设正方形边长为a,则AO=√2/2 a,∴A1O=√3/√2 a,

∵OC=√2/2 a,MC=a/2 ∴OM=√3/2 a,∵A1C1=√2a,C1M=a/2,∴A1M=3/2

△A1OC1,由余弦定理得cos∠A1OM=(A1O²+OM²-A1M²) /  (2A1O*OM)=0

即所求角的余弦值为0。