Question

如圖,已知直四棱柱ABCD—A₁B₁C₁D₁中,底面ABCD是直角梯形,∠DAB

=90°,AB∥DC,AB=2,AD=DC=1,AA₁=2,E為BC₁的中點.

(1)求證:AB₁⊥BC₁;

(2)若F是棱DD₁上的一點,當的值為多少時,能使二面角F-AC-E為直二面角?請給出證明.

Answer 1

解法一:(1)證明:建立如圖所示的空間直角坐標系A—xyz,則依題設知

A(0,0,0),B(0,2,0),B₁(0,2,),C₁(1,1,),                                                       

∴=(0,2,),=(1,-1,).                                                                ?

∴·=(0,2,)·(1,-1,)?

=0×1+2×(-1)+ ×=0.                                                                                    ?

故AB₁⊥BC₁.                                                                                                          ?

(2)設F(1,0,A),∵C(1,1,0),E(,,),?

∴=(1,1,0),=(1,-1,0).?

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

=(1,1,0)-(1,0,A)=(0,1,-A),?

·=(1,1,0)·(1,-1,0)=0,·=(1,1,0)·(-,,)=0,?

∴AC⊥BC,AC⊥CE.                                                                                                ?

若使二面角F-AC-E為直二面角,只需EC⊥FC即可,這樣面FAC⊥面ACE.?

∵·=(0,1,-A)·(-,,)=-A,?

當·=0時,得A=,即F(1,0,),                                                      ?

故當=1時,二面角F-AC-E為直二面角.                                                           ?

解法二:(1)證明:∵AB=2,AC==,∠CAB=45°,                                 ?

易知∠ABC=45°,∴∠ACB=90°,故AC⊥面CBB₁C₁.??

又∵AA₁=,BC=ABsin45°=,?

故四邊形CBB₁C₁為正方形.連結(jié)EB₁,則B₁C⊥C₁B.                                                  ?

由三垂線定理可得,AB₁⊥BC₁.                                                                                 ?

(2)∵AC⊥面CBB₁C₁,?

∴B₁C⊥AC.若使二面角F-AC-E為直二面角,只需B₁C⊥FC即可,這樣面FAC⊥面ACB₁.

?

連結(jié)A₁B,A₁C₁,AC∥A₁C₁,∴A₁C₁⊥面CBB₁C₁.由三垂線定理可得,A₁B⊥CB₁,當F為棱DD₁的中點時,易證FC∥A₁B,此時有CB₁⊥FC,故當=1時,二面角F-AC-E為直二面角.