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如图,在三棱柱ABC

解答:证明:(1)∵△ABC为正三角形,D是BC的中点∴BC⊥AD,…(1分)∵AA1⊥平面ABC,BC?平面ABC,∴BC⊥AA1 …(3分)∵AD,AA1是平面DAA1内的两条相交直线,∴BC⊥平面DAA1 …(5分)∵A1D?平面DAA1∴BC⊥A1D …(6分)(2)∵D,E,F分别为BC,B1C1,A1B1的中...

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:(1)设O为AC的中点,连接A1O,A1C,∵AA1=AC,∠A1AC=60°,∴△A1AC为正三角形,∴A1O⊥AC∵平面A1ACC1⊥平面ABC,平面A1ACC1∩平面ABC=AC,A1O?平面A1ACC1,∴A1O⊥平面ABC,∵△ABC为正三角形,∴OB⊥AC建立如图所示的空间直角坐标系,则A(1,0,0),B...

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...

证明:(1)∵ABC-A1B1C1为三棱柱,D是BC中点,AA1⊥平面ABC,AD?平面ABC,∴AA1⊥AD;又AA1∥BB1,∴AD⊥BB1;又底面ABC为正三角形,D是BC中点,∴AD⊥BC,而BC∩BB1=B,∴AD⊥平面BCC1B1,BC1?平面BCC1B1,∴AD⊥BC1;(2))取C1B1的中点E,连接A1E,ED,则...

(Ⅰ)取AB的中点为M,连接EF,EM,CM,∵E是A1B的中点,F是棱CC1中点,∴EM∥AA1,FC∥AA1,EM=FC=12AA1,则四边形EMCF是平行四边形,∴EF∥CM,又∵△ABC为正三角形,侧面AA1C1C是正方形,∴AA1=AB,∴AE⊥A1B,CM⊥AB,∵侧棱AA1⊥平面ABC,∴CM⊥AA1,∴CM⊥...

如图所示:取BC的中点O,连接OA,OA1.在直三棱柱ABC-A1B1C1中,AA1⊥底面ABC,∵△ABC是边长为3正三角形,点O是BC的中点,∴OA=3×32=32,BC⊥OA,由三垂线定理可得:BC⊥A1O.∴∠AOA1是二面角A1-BC-A的平面角.在Rt△OAA1中,由OA=AA1=32可得∠AOA1=4...

解:如图所示

解答:(Ⅰ)证明:∵AA1⊥平面A1B1C1,∴AA1⊥A1C1.又 A1C1⊥A1B1,AA1∩A1B1=A1∴A1C1⊥平面AA1B1B.∴A1C1⊥AB1又四边形AA1B1B是正方形,AB1⊥A1B,A1B∩A1C1=A1∴AB1⊥平面A1BC1.(Ⅱ)设AB1∩BA1=O,连结AC1,∵AB=AC=AA1=a,A1C1⊥A1B1,AA1⊥平面A1B1C1,∴△...

解:(Ⅰ)取AB中点E,连接DE,CE因为直棱柱,CC1⊥面ABC,所以CC1⊥AB,又因为△ABC为等腰直角三角形,所以CE⊥AB,所以AB⊥面DEC,即AB⊥DE,所以∠DEC即为二面角D-AB-C的平面角因为CD=1,CE=2,则tan∠DEC=DCCE=12=22(II)连接BC1.因为直棱柱,所...

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