www.1862.net > (2014?上饶一模)如图,在斜三棱柱ABC

(2014?上饶一模)如图,在斜三棱柱ABC

(1)延长B1E交BC于F,∵△B1EC1∽△FEB,BE=12EC1∴BF=12B1C1=12BC,从而F为BC的中点. (2分)∵G为△ABC的重心,∴A、G、F三点共线,且=FGFA=FEFB1=13,∴GE∥AB1,又GE?侧面AA1B1B,AB1?侧面AA1B1B,∴GE∥侧面AA1B1B (4分)(2)在侧面AA1B1B内,过B...

解答:(1)解:∵AB⊥侧面BB1C1C,且AB∥A1B1,∴四棱锥的高h=AB=1 …(2分)又S底面=2?12CB?CC1?sin∠C1CB=3…(4分)∴四棱锥的体积为13?3?1=33…(6分)(2)证明:在△BCC1中,∵BC=1,CC1=2,∠BCC1=π3,∴BC1=1+4?2?1?2?12=3,∴∠CBC1=90°,∴BC⊥BC1,...

(Ⅰ)B1C与AC1垂直.证明如下:连接BC1,由题意知B1C⊥BC1作B1D⊥AB,由条件知B1D⊥面ABC,又侧棱与底面所成的角为60°,∴D为AB的中点,∴CD⊥AB,CD为CB1在底面ABC上的射影又B1C⊥AB,∴B1C⊥面ABC1,∴B1C⊥AC1(Ⅱ)由题意知cos∠B1BC=cos∠B1BA?cos∠CBA=...

设三棱柱ABC-A1B1C1的体积为V∵侧棱AA1和BB1上各有一动点P,Q满足A1P=BQ,∴四边形PQBA与四边形PQB1A1的面积相等,∵M是棱CA上的动点,∴M是C时,VM?ABQPVABC?A1B1C1?VM?ABQP 最大又四棱椎C-PQBA的体积等于三棱锥C-ABA1的体积等于13V,∴VM?ABQPVABC?...

证明:(1)∵AC2+BC2=AB2,∴AC⊥BC.又∵C1C∥AA1,AA1⊥底面ABC,∴C1C⊥底面ABC,∴AC⊥CC1 .又BC∩CC1=C,∴AC⊥平面BCC1B1 .而BC1?平面BCC1B1,∴AC⊥BC1 .(2)设BC1∩B1C=O,则O为BC1的中点,连接OD,∵D为AB的中点,∴OD∥AC1,又∵OD?平面CDB1,AC1?平...

证明:(1)∵A1B⊥面ABC,∴A1B⊥AC,------(1分)又AB⊥AC,AB∩A1B=B∴AC⊥面AB1B,------(3分)∵AC?面A1AC,∴平面A1AC⊥平面AB1B;------(4分)(2)如图,以A为原点建立空间直角坐标系,则C(2,0,0),B(02,0),A1(0,2,2),B1(0,4,2...

∵AC⊥AB,AC⊥BC1,∴AC⊥平面ABC1,AC?平面ABC,∴平面ABC1⊥平面ABC,∴C1在平面ABC上的射影H必在两平面的交线AB上.故答案为:AB

(I)证明:在直三棱柱ABC-A1B1C1中,有AA1⊥平面ABC.∴AA1⊥AC,又AA1=AC,∴A1C⊥AC1. …(2分)又BC1⊥A1C,∴A1C⊥平面ABC1,∵A1C?平面A1C1CA,∴平面ABC1⊥平面A1C1CA. …(4分)(II)解:取AA1中点F,连EF,FD,当E为B1B中点时,EF∥AB,DF∥AC1.即...

(1)证明:∵斜三棱柱ABC-A1B1C1中,点A1在底面ABC上的射影恰好是AB的中点O,底面ABC是正三角形,其重心为G点,D是BC中点,B1D交BC1于E,∴DEEB1=BDB1C1=12,连结AB1,则DEEB1=DGGA=12,∴GE∥AB1,∵GE不包含于侧面AA1B1B,AB1?侧面AA1B1B,∴GE∥...

(1)证明:∵侧面AA1B1B⊥底面ABC,侧棱AA1与底面ABC成60°的角,∴∠A1AB=60°,又AA1=AB=2,取AB的中点O,则AO⊥底面ABC.以O为原点建立空间直角坐标系O-xyz如图,则A(0,-1,0),B(0,1,0),C(3,0,0),A1(0,0,3),B1(0,2,3),C1(3,1...

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