P,Q,R分别为正方体ABCD-A1B1C1D1的棱AB,BB1,BC的中点.求证:BD1⊥平面PQR.

∵ABCD－A1B1C1D1是正方体,∴ABCD是正方形,∴AC⊥BD.
∵ABCD－A1B1C1D1是正方体,∴BB11⊥平面ABCD,∴AC⊥BB1.
由AC⊥BD、AC⊥BB1,得：AC⊥平面BB1D1D,∴BD1⊥AC.
∵P、R分别是AB、BC的中点,∴PR∥AC,∴BD1⊥PR.
∵ABCD－A1B1C1D1是正方体,∴AA1B1B是正方形,∴AB1⊥A1B.
∵ABCD－A1B1C1D1是正方体,∴BC⊥平面AA1B1B,∴AB1⊥BC.
由AB1⊥A1B、AB1⊥BC,得：AB1⊥平面A1BCD1,∴BD1⊥AB1.
∵P、Q分别是AB、BB1的中点,∴PQ∥AB1,∴BD1⊥PQ.
由BD1⊥PR、BD1⊥PQ,得：BD1⊥平面PQR.