题目

20.已知三棱柱ABC－A1B1C1中底面边长和侧棱长均为a,侧面A1ACC1⊥底面ABC,A1B＝a.20题图(Ⅰ)求异面直线AC与BC1所成角的余弦值；(Ⅱ)求证A1B⊥面AB1C. 答案：  解:过点B作BO⊥AC,垂足为点O,则BO⊥侧面ACC1A1,连结A1O,  在Rt△A1BO中,A1B＝a,BO＝a,∴A1O＝a,又AA1＝a,AO＝.∴△A1AO为直角三角形,A1O⊥AC,A1O⊥底面ABC.解法一:(Ⅰ)∵　A1C1∥AC,∴　∠BC1A1为异面直线AC与BC1所成的角.∵　A1O⊥面ABC,AC⊥BO,∴　AC⊥A1B,∴ A1C1⊥A1B.在Rt△A1BC1中,A1B＝a,A1C1＝a,∴　BC1＝a∴cosBC1A1＝,所以,异面直线AC与BC1所成角的余弦值为.(Ⅱ)设A1B与AB1相交于点D,∵　ABB1A1为菱形,∴　AB1⊥A1B.又　A1B⊥AC, AB1与AC是平面AB1C内两条相交直线,所以A1B⊥面AB1C.解法二:(Ⅰ)如图,建立坐标系,原点为BO⊥AC的垂足O.由题设条件可得B(a,0,0),C1(0,a,a),A(0,－a,0),C(0,a,0),∴  ＝(－a,a,a), ＝(0,a,0).设与的夹角为θ,则cosθ＝＝＝,所以,异面直线AC与BC1所成角的余弦值为.(Ⅱ)A1(0,0,a),B(,0,0),∴　＝(a,0,－a),    ＝(0,a,0),·＝0.∴  A1B⊥AC1.又ABB1A1为菱形,∴　A1B⊥AB1.又因为AB1与AC为平面AB1C内两条相交直线,　　所以A1B⊥平面AB1C.

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