b/(3^0.5)
過(guò)A1做平面ABC的垂線,垂足為O,過(guò)O作OD垂直于AB,連接A1D.
∵OA1⊥平面ABC,∴OA1⊥AB
∵OD⊥AB,∴AB⊥平面ODA1
∴AB⊥DA1,∴AD=DA1=b/(2^0.5)
∵O在角BAC的平分線上,∴角OAB=30°
∴OD=AD/(3^0.5)=b/(6^0.5)
由勾股定理OA1=b/(3^0.5)