Question

如圖,矩形ABCD和AB′C′D全等,且所在平面所成的二面角為α,記兩個矩形對角線的交點分別為Q,Q′,AB=a,AD=b.

(1)求證:QQ′∥平面ABB′;

(2)當b=a,且α=時,求異面直線AC與DB′所成的角;

(3)當a＞b,且AC⊥DB′時,求二面角α的余弦值(用a,b表示).

Answer 1

解:(1)證明:連結BB′,∵Q,Q′分別是BD,B′D′的中點,

∴QQ′∥BB′.而BB′平面ABB′,∴QQ′∥平面ABB′.

(2)以A為原點,AB,AD分別為x軸,z軸建立空間直角坐標系,如圖:

由條件可設A(0,0,0),B(a,0,0),C(a,0,b),D(0,0,b),

又∠BAB′=,AB′=a,

∴B′(,,0),C′(,,b),

=(a,0,b),=(,,-b).設異面直線AC與DB′所成角為θ,

cosθ====,所以異面直線AC與DB′所成角為60°.

(3)設B′(p,q,0),C′(p,q,b),∵AB′=a,∴p²+q²=a².

∴=(p,q,-b).又有=(a,0,b),∵AC⊥DB′,

∴·=pa-b²=0,得pa=b².設平面AB′C′D的法向量為n=(x,y,z),

∵n⊥,n⊥,而=(0,0,b),=(p,q,0),

∴n=(,1,0),設平面ABCD的法向量為m,則m=(0,±1,0),

∴cosα=±=±=±.