www.1862.net > (2011?商丘二模)如图,三棱柱ABC

(2011?商丘二模)如图,三棱柱ABC

解答:解:(1)设AB1与A1B相交于点P,连接PD,则P为AB1中点,∵D为AC中点,∴PD∥B1C.又∵PD∥平面A1BD,∴B1C∥平面A1BD.(2)∵正三棱住ABC-A1B1C1,∴AA1⊥底面ABC.又∵BD⊥AC∴A1D⊥BD∴∠A1DA就是二面角A1-BD-A的平面角.∵AA1=3,AD=12AC=1∴tan∠A1DA=A1...

解答:(本小题满分12分)解:(I)取AB中点O,连接OM,OC.∵M为A1B1中点,∴MO∥A1A,又A1A⊥平面ABC,∴MO⊥平面ABC,∴MO⊥AB…(2分)∵△ABC为正三角形,∴AB⊥CO 又MO∩CO=O,∴AB⊥平面OMC又∵MC?平面OMC∴AB⊥MC…(5分)(Ⅱ)如图,VA1-ABP=VP-A1BA=13S△A1...

解法一:(Ⅰ)证明:连接AO,∵A1O⊥面ABC,BC?面ABC∴A1O⊥BC∵AO⊥BC,A1O∩AO=O∴BC⊥平面A1OA∵A1A?平面A1OA∴A1A⊥BC.…3分(Ⅱ)解:由(Ⅰ)得∠A1AO=45°由底面是边长为23的正三角形,可知AO=3,∴A1O=3,AA1=32过O作OE⊥AC于E,连接A1E,则∠A1EO为二面角A...

(1)证明:∵正三棱住ABC-A1B1C1,∴AA1⊥底面ABC,又∵BD⊥AC,A1A∩AC=A,∴BD⊥平面A1ACC1,又∵BD?平面A1BD,∴平面A1BD⊥平面A1ACC1…6分(2)解:作AM⊥A1D,M为垂足,由(1)知AM⊥平面A1DB,设AB1与A1B相交于点P,连接MP,则∠APM就是直线A1B与平面A1B...

解:(Ⅰ)E为BC1中点.(2分)因为A1A=A1C,且O为d的中点,所以A1O⊥AC.又由题意可知,平面AA1C1C⊥平面ABC,交线为AC,且A1O?平面AA1C1C,所以A1O⊥平面ABC.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.(1分)由题意可...

解:(Ⅰ)过A1作A1H⊥平面ABC,垂足为H.过H作HD⊥AB,连A1D则A1D⊥AB作HF⊥AC,连结A1F则A1F⊥AC,又∠A1AB=∠A1AC=45°∴Rt△DAA1≌Rt△FAA1,∴AD=AF∴Rt△ADH≌Rt△FAH所以H在∠CAB平分线AE上,由△ABC为正三角形,∴BC⊥AE?BC⊥AA1异面直线AA1与BC所成角为90°;--...

∵三棱柱ABC-A1B1C1是正三棱柱,∴△ABC是等边三角形,且四边形BCB1C1是平行四边形∴BC∥B1C1,可得∠ACB(或其补角)就是异面直线AC与B1C1所成的角∵等边△ABC中,∠ACB=60°∴异面直线AC与B1C1所成的角等于60°故答案为:60°

三棱柱外接球的半径为:r,43πr3=32π3所以r=2,即OA=2,所以三棱柱的高为:222?(23×32×3) 2=23,三棱柱的体积:sh=34×(3)2×23=92.故答案为:92.

解答:解:(1)如图所示,以C为原点,CA、CB、CC1为坐标轴,建立空间直角坐标系C-xyz.则C(0,0,0),A(2,0,0),B(0,2,0),C1(0,0,2),B1(0,2,2),D(2,0,1).所以DC1=(-2,0,1),B1C=(0,-2,-2). 所以cos<DC1,B...

解答:证明:(1)∵AB=AC,D为BC的中点∵E为AB的中点,连接CE交AD于O,连接FO,∴COCE=CFCC1=23∴FO∥EC1(2分)∵FO?平面AFD,C1E?平面AFD(4分)∴C1E∥平面AFD(5分)(2)在平面C1CBB1内,过C作CG⊥DF,交BB1于G在△RtFCD 和△RtCBG中FC=CB,∠CFD=∠BC...

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