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已知斜三棱柱ABC

解:(I)取AC中点D,连接A1D,则A1D⊥AC.又∵侧面ACC1A1与底面ABC垂直,交线为AC,∵A1D⊥面ABC(2分)∴A1D⊥BC.假设AA1与平面A1BC垂直,则A1A⊥BC.又A1D⊥BC,由线面垂直的判定定理,BC⊥面A1AC,所以BC⊥AC,这样在△ABC中有两个直角,与三角形内角...

解法一:如图建立空间直角坐标系,(1)有条件知B(0,0,0),C(0,2,0),A(22,0,0),(1分)由面ACC1A1⊥面ABC,AA1⊥A1C,AA1=A1C,知A1(2,1,3)(2分)AA1=(?2,1,3),BC=(0,2,0),∵AA1?

(1)解:如图作A1D⊥AC,垂足为D,由面A1ACC1⊥面ABC,得A1D⊥面ABC,所以∠A1AD为A1A与面ABC所成的角.因为AA1⊥A1C,AA1=A1C,所以∠A1AD=45°为所求.(2)解:作DE⊥AB,垂足为E,连A1E,则由A1D⊥面ABC,得A1E⊥AB.所以∠A1ED是面A1ABB1与面ABC所成...

解答:解:(1)因为侧面A1ACC1⊥底面ABC,AA1?侧面A1ACC1,侧面A1ACC1∩底面ABC=AC所以直线AA1在底面ABC内的射影为直线AC故∠A1AC为侧棱AA1与底面ABC所成的角又AC=AA1=A1C,所以∠A1AC=60°为所求. (4分)(2)取AC,AB的中点分别为M,N,连结A1M...

取AA1的中点E,连接CE、BE。易知CB垂直于面AA1C1C,所以AA1垂直于CB,结合AA1垂直于CE知,AA1垂直于面CEB,在直角三角形CEB中斜边BE上的高即为所求。

解答:(1)证明:取AC中点P,则BP⊥AC∵平面A1ACC1⊥平面ABC,平面A1ACC1∩平面ABC=AC,∴BP⊥平面A1ACC1,∵A1C?平面A1ACC1,∴A1C⊥BP∵A1C⊥AC1,AC1∥PM∴A1C⊥PM∵BP∩PM=P∴A1C⊥面BPM∵BM?面BPM∴A1C⊥BM;(2)解:作PQ⊥A1A于Q,连接BQ∵BP⊥平面A1ACC1,∴A1A⊥...

证明:(1)∵AB⊥AC,面ACC1A1⊥面ABC,∴AB⊥面ACC1A1,即有AB⊥CD;又AC=A1C,D为AA1中点,则CD⊥AA1∴CD⊥面ABB1A1 (2)第二问好多打不出来,我说说吧。。。 (2)建立空间直角坐标系C-xyz,由=λ,可得E点坐标为((1-λ)a,a,λa).求出面A1C1A地...

解答:(Ⅰ)解:连接A1B交AB1于Q,则Q为A1B中点,连接PQ,∵P是BC的中点,∴PQ∥A1C.…(4分)∵PQ?平面AB1P,A1C?平面AB1P,∴A1C∥平面AB1P. …(6分)(Ⅱ)取A1C1中点M,连B1M、AM,则B1M⊥A1C1.∵平面ACC1A1⊥平面ABC,∴平面ACC1A1⊥平面A1B1C1.∴B1M...

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:取AC的中点E,连接BE,C1E,∵正三棱柱ABC-A1B1C1中,∴BE⊥面ACC1A1,∴∠BC1E就是BC1与侧面ACC1A1所成的角,BC1=3,BE=32,∴sinθ=12,θ=30°.故答案为30°.

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