题目

18<span 答案:方法一 (1)证:∵CD⊥AB,CD⊥BC,∴CD⊥平面ABC                                                        又∵CDÌ平面ACD,∴平面ACD⊥平面ABC   (2)解:∵AB⊥BC,AB⊥CD,∴AB⊥平面BCD,故AB⊥BD ∴∠CBD是二面角C-AB-D的平面角          ∵在Rt△BCD中,BC = CD,∴∠CBD = 45° 即二面角C-AB-D的大小为45°             (3)解:过点B作BH⊥AC,垂足为H,连结DH ∵平面ACD⊥平面ABC,∴BH⊥平面ACD, ∴∠BDH为BD与平面ACD所成的角          设AB = a,在Rt△BHD中,, ∴ 又,∴                                                                                        方法二 (1)同方法一                                                                                                                (2)解:设以过B点且∥CD的向量为x轴,为y轴和z轴建立如图所示的空间直角坐标系,设AB = a,则A(0,0,a),C(0,1,0),D(1,1,0), = (1,1,0), = (0,0,a) 平面ABC的法向量 = (1,0,0) 设平面ABD的一个法向量为n = (x,y,z),则 取n = (1,-1,0)                           6分 ∴二面角C-AB-D的大小为45°                                                                            (3)解: = (0,1,-a), = (1,0,0), = (1,1,0) 设平面ACD的一个法向量是m = (x,y,z),则 ∴可取m = (0,a,1),设直线BD与平面ACD所成角为,则向量、m的夹角为 故                                                                        即 又,∴                                                                                       
数学 试题推荐