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1)Mgtanθ答:所施加水平恒力的大小F为(2?1)Mgtanθ.
(2014?奉贤区二模)如图,在直三棱柱ABC-A1B1C1中,∠BAC=90°,AB=AC=...(理)解法一:如图1以A1为原点,A1B1所在直线为x轴,A1C1所在直线为y轴,A1A所在直线为z轴建系,则A(0,0,1),D(12,12,0),则AD=(12,12,?1),(2分)设平面A1BC1的一个法向量n=(x,y,z),∵A1 C1=(0,1,0),A1B=(1,0,1),∴n?
(2014?宜昌二模)如图所示,在直三棱柱ABC-A1B1C1中,AB=BB1,AC1⊥平面A...∴AC1⊥A1B,∴A1B⊥面AB1C1,∴A1B⊥B1C1,(4分)又在直棱柱ABC-A1B1C1中,BB1⊥B1C1,∴B1C1 ⊥平面ABB1A 1 .(5分)(Ⅱ)解:设AB=BB1=a,CE=x,∵D为AC的中点,且AC1⊥A1D,
(2014?蚌埠二模)如图,正三棱柱ABC-A1B1C1中,AB=2,AA1=3,D为C1B的中点...解:(I)连接DP、AC1,∵△ABC1中,P、D分别为AB、BC1中点∴DP∥AC1,∵AC1?平面ACC1A1,DP?平面ACC1A1,∴DP∥平面ACC1A1(II)由AP=3PB,得PB=14AB=12过点D作DE⊥BC于E,则DE∥CC1且DE=12CC1又∵CC1⊥平面ABC,∴DE⊥平面BCP∵CC1=3,∴DE=32∵S△BCP=12×2×12×sin60°=...
(2014?仁寿县模拟)如图,三棱柱ABC-A1B1C1中,CA=CB,AB=AA1,∠BAA1=60...(Ⅰ)取AB的中点O,连接OC,OA1,A1B,因为CA=CB,所以OC⊥AB,由于AB=AA1,∠BAA1=60°,所以△AA1B为等边三角形,所以OA1⊥AB,又因为OC∩OA1=O,所以AB⊥平面OA1C,又A1C?平面OA1C,故AB⊥A1C;(Ⅱ)由(Ⅰ)知OC⊥AB,OA1⊥AB,又平面ABC⊥平面AA1B1B,交线为AB,所以OC⊥...
(2014?南京模拟)如图,在斜三棱柱ABC-A1B1C1中,侧面A1ACC1是边长为2的...证明:(1)由题意,平面ABC∥平面A1B1C1,又∵平面A1B1M与平面ABC交于直线MN,与平面A1B1C1交于直线A1B1,∴MN∥A1B1.∵AB∥A1B1,∴MN∥AB,∴CNAN=CMBM.∵M为AB的中点,∴CNAN=1,∴N为AC中点.(2)∵四边形A1ACC1是边长为2的菱形,∠A1AC=60°.在三角形A1AN中,AN=1,AA1=2,...
(2014?包头二模)如图,直三棱柱ABC-A1B1C1中AA1=2AC=2BC,D是AA1的中点...(1)证明:由题意知直三棱柱ABC-A1B1C1的侧面为矩形,∵D是AA1的中点,∴DC=DC1,又AA1=2A1C1,∴DC12+DC2=CC12,∴CD⊥DC1,而CD⊥B1D,B1D∩C1D=D,∴CD⊥平面B1C1D,∵B1C1?平面B1C1D,∴CD⊥B1C1.(2)解:由(1)知B1C1⊥CD,且B1C1⊥C1C,∴B1C1⊥平面ACC1A1,∴...
(2014?商丘二模)已知直三棱柱ABC-A1B1C1中,AC=BC,点D是AB的中点.(1...面CA1D,BC1?面CA1D,∴BC1∥平面CA1D;(2)AC=BC,D是AB的中点,∴AB⊥CD,又∵AA1⊥面ABC,CD?面ABC,∴AA1⊥CD,∵AA1∩AB=A,∴CD⊥面AA1B1B,又∵CD?面CA1D,∴平面CA1D⊥平面AA1B1B(3)则由(2)知CD⊥面ABB1B,∴三棱锥B1-A1DC底面B1A1D上的高就是CD=3,又∵BD=...
(2014?甘肃二模)如图,直三棱柱ABC-A1B1C1中,AC⊥AB,AB=2AA1,M是AB的...(6分)(2)连结B1M,…(7分)因为三棱柱ABC-A1B1C1为直三棱柱,∴AA1⊥平面ABC,∴AA1⊥AB,即四边形ABB1A1为矩形,且AB=2AA1,∵M是AB的中点,∴B1M⊥A1M,又A1C1⊥平面ABB1A1,∴A1C1⊥B1M,从而B1M⊥平面A1MC1,…(9分)∴MC1是B1C1在平面A1MC1内的射影,∴B1C1与平面A1MC1...
(2014?广州模拟)如图,三棱柱ABC-A1B1C1的侧棱AA1⊥平面ABC,△ABC为正...32x=33,∴x3=8,x=2(Ⅱ)解法一:将侧面BCC1B1展开到侧面A1ACC1得到矩形ABB1A1,连结A1B,交C1C于点F,此时点F使得A1F+BF最小.此时FC平行且等于A1A的一半,∴F为C1C的中点.取AB中点O,连接OE,EF,OC,∴OEFC为平行四边形,∵△ABC为正三角形,∴OC⊥AB,又AA1⊥平面ABC,∴OC⊥...