如图,把B1BCC1向后延伸一倍,M是C1Q的中点,则AE‖＝CM.
设AB＝2,则CM＝√5,CF＝√6,MF＝√5, 从余弦定理cos∠MCF＝3/√30
异面直线AE与CF所成角＝arccos3/√30≈56°47′21〃