如图,在三棱柱ABC-A1B1C1中,侧面AA1C1C⊥底面ABC,AA1=A1C=AC=2,AB=BC,且AB⊥BC,O为AC中点.(1)求 如图,三棱柱ABC-A1B1C1中,侧面AA1CC1⊥底面A...
所以A1O⊥平面ABC.…(3分)
如图,以O为原点,OB,OC,OA1所在直线分别为x,y,z轴建立空间直角坐标系.由题意可知A1A=A1C=AC=2,又AB=BC,AB⊥BC,∴OB=
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所以得:O(0,0,0),A(0,?1,0),A1(0,0,
(2013?长春一模)如图,在三棱柱ABC-A1B1C1中,侧面AA1C1C⊥底面ABC,AA1=A1C=AC=2,AB=BC,AB⊥BC,O为~ 证明:(1)∵AA1=A1C=AC=2,且O为AC中点,∴A1O⊥AC,又∵侧面AA1C1C⊥底面ABC,侧面AA1C1C∩底面ABC=AC,A1O?侧面AA1C1C,∴A1O⊥平面ABC.(6分)解:(2)VE?BCC1=112VABC?A1B1C1=14VA1?BCC1,因此BE=14BA1,即A1E=34A1B,又在Rt△A1OB中,A1O⊥OB,A1O=3,BO=1可得A1B=2,则A1E的长度为32.(12分) 证明:(1)∵A1A=A1C,且O为AC的中点,∴A1O⊥AC.又侧面AA1C1C⊥底面ABC,其交线为AC,且A1O∈平面AA1C1C,所以A1O⊥底面ABC.…..(2分)以O为坐标原点,OB,OC,OA1所在直线分别为x,y,z轴建立空间直角坐标系.由已知可得:O(0,0,0),A(0,-1,0),A1(0,0,3),C(0,1,0),C1(0,2,3),B(1,0,0),E(12,1,32).则有:A1C=(0,1,?3),AA1=(0,1,3),<div style="background-image: url(http://hiphotos.baidu.com/zhidao/pic/item/a2cc7cd98d1001e9cd3ea5bbbb0e7bec54e79700.jpg); background-attachment: initial; background-origin: initial; background-clip: initial; background-color: initial; height: 5px; float: left; overflow-x: hidden; overflow-y: hidden; width: 11px; background-position: initial initial; background-repeat: repe 如图所示,在正三棱柱ABC—A1B1C1中,AB=2,AA1=2,由定点B沿棱柱侧面... 如右图,在三棱柱ABC—A1B1C1中,侧棱AA1⊥底面ABC,AB=AC=1,AA1=2,∠B... 如图,在三棱柱ABC-A1B1C1中,侧面ABB1A1,ACC1A1均为正方形,∠BAC=90°... 如图所示,在直三棱柱ABC-A1B1C1中,∠ACB=90° 如图,在三棱柱ABC-A1B1C1中,底面是边长为3的正三角形,侧棱长为3,且侧 ... 如图所示,在三棱柱ABC—A1B1C1中,A1A⊥平面ABC,AB=2BC,AC=AA1=3^1... (2009?济宁一模)如图,在三棱柱ABC-A1B1C1中,所有的棱长都为2,∠A1AC... (2012•重庆)如图,在直三棱柱ABC-A1B1C1中,AB=4,AC=BC=3,D为AB的... 如图,在正三棱柱ABC-A1B1C1中,AB=3,AA1=6,M为AA1上的点,且AM=2MA1,P... 如图,在三棱柱ABC-A1B1C1中,AB⊥侧面BB1C1C,已知AB=1,BC=1,BB1=2... |