Question

已知四棱锥P—ABCD的底面为直角梯形,AB∥DC,∠DAB=90°,PA⊥底面ABCD,且PA=AD=DC=AB=1,PM=kPB.

(1)若k=时,试判断PD与平面AMC的位置关系,并给予证明;

(2)k为何值时,二面角A-MC-B大小为?

Answer 1

解:(1)PD∥平面AMC.?

以AD,AB,AP为x,y,z轴建立如图坐标系,则P(0,0,1),C(1,1,0),B(0,2,0),D(1,0,0),M(0, ,).                                                                                                                   ?

∴=(-1,0,1),=(0,,),=(-1,-,).                                          ?

=λ+μ,?

∴ ?

∴λ=,μ=1.                                                                                                     ?

=+,∴DP∥平面AMC.                                                                   ?

(2)设平面AMC的法向量n₁=(a,b,1),平面MCB的法向量为n₂=(c,d,1).?

M(0,2k,1-k),=(1,1,0),=(0,2k,1-k),=(1,-1,0),=(0,2,-1).?

∴∴∴

∴n₁=(,,1).                                                                                     ?

∴∴∴∴n₂=(,,1).                              ?

若二面角AMCB的大小为,?

则cos〈n₁,n₂〉==,?

∴(+1)×=2.?

∴=.?

∴k²-6k+3=0.?

∴k=3±(∵0≤k≤1,∴3+6舍去).?

∴k=3-，?

即当k=3-时,二面角AMCB的大小为.