(Ⅰ)连结CP并延长与DA的延长线交于M点,证明BC∥AD,PQ∥MD1,又MD1⊂平面A1D1DA,PQ⊄平面A1D1DA,证明PQ∥平面A1D1DA;
(Ⅱ)R是AB上的点,当的值为时,能使平面PQR∥平面A1D1DA,通过证明PR∥平面A1D1DA,又PQ∩PR=P,PQ∥平面A1D1DA.然后证明即可.
(Ⅰ)证明:连结CP并延长与DA的延长线交于M点,
因为四边形ABCD为正方形,所以BC∥AD,
故△PBC∽△PDM,所以,
又因为,所以,所以PQ∥MD1.
又MD1⊂平面A1D1DA,PQ⊄平面A1D1DA,故PQ∥平面A1D1DA. …(6分)
(Ⅱ)当的值为时,能使平面PQR∥平面A1D1DA.
证明:因为,即有,故,所以PR∥DA.
又DA⊂平面A1D1DA,PR⊄平面A1D1DA,
所以PR∥平面A1D1DA,又PQ∩PR=P,PQ∥平面A1D1DA.
所以平面PQR∥平面A1D1DA.…(12分)