参考答案：一单选题：12345678CBACDDAB二．多项选择题：91011BCDABDACD三填空题：12.2513.14.-1265四解答题：5【解析】（）记，15.1ABa,ADb,AA1c因为平行六面体ABCDA1B1C1D1的各条棱长均为2，A1ABA1ADBAD60，rrr所以abc2，abacbc22cos602，.............................................2...................................................................................................................43�1�=�−5�−�..............................................................................................6322341（��）记=(�−5�−�)=5，2ABa,ADb,AA1c因为平行六面体ABCDA1B1C1D1的各条棱长均为2，A1ABA1ADBAD60，rrr所以abc2，abacbc22cos602，因为AC1ABBCCC1ABADAA1abc，A1BABAA1ac,BDADABba，..................................................8所以22，AC1A1Babcacaabacacbcc4222240.........11同理AC1BD0，则AC1A1B,AC1BD，又A1BBDB,A1B,BD平面A1BD，所以AC1平面A1BD..........................1316（1）首先根据圆的弦长公式设圆心到直线距离为d,可得,所以...........................................................................222若直线M斜N率=不2存R在−时d，=直2线方30程为x=d2=此时6d=2不满足题意。.................................................3{#{QQABaQQAggCIAAIAABgCEwUCCEGQkhAAASgOBAAIsAAByBFABAA=}#}所以直线斜率存在，设直线方程为y-3=k(x-2)圆心到直线距离为,2−2k2所以即得直线斜率d=k+1=...6...............................................................................................6所学直线方程为k=−2±3或.........................7(2)设圆心到直线yA=B(的−距2离+为3)x，+圆7心−到2直3线CyD=的(−距2离−为3),x+7+23.................�..1.................................................�..2.................................1022212�+�=