在三棱柱中ABC-A1B1C1中 AA1垂直平面ABC AB垂直面BCC1B1 且AB=BC=BB...答:如图,A(1,0,0),C(0,1,0),B1(0,0,1),A1(1,0,1),C1(0,1,1) M(1/2,0,0),N(1/2,1/2,1/2)A1B1C方程为y+z = 1, 法向量(0,1,1)MN(0,1/2,1/2)与(0,1,1)平行,所以MN与A1B1C垂直 A1B1=1, B1C = 根号(2),A1C=根号(3),A1B1C的面积 = 根号(2)...
如图,在直三棱柱ABC–A1B1C1中,AB垂直AC,AB=AC=1,AA1=2,D、E分别是BB...答:∵三棱柱是直三棱柱,∴MH⊥面A1B1C1 ∴A1M在面A1B1C1上的射影是A1H 易证H是B1C1中点,而A1B1=A1C1 ∴A1H⊥B1C1,∴A1M⊥B1C1 ∵DE∥B1C1,∴DE⊥A1M 勾股定理得AD=AE=DE=AC=√2 ∴△ADE是等边三角形,而M是DE中点 ∴DE⊥AM ∴DE⊥面AA1M,∴面ADE⊥面AA1M (2)连接BC1,那麼M是...
如图,已知△ABC∽△A1B1C1,相似比为K(K>1)且△ABC的三边长分别为a,b...答:解:(1)证:Q△ ABC ∽△ A1 B1C1 ,且相似比为 k ( k > 1), a = k, a = ka1. ∴ a1 又Q c = a1, a = kc.(2)解:取 a = 8,b = 6,c = 4,同时取a1 = 4,b1 = 3,c1 = 2.此时 a b c = = = 2,△ ABC ∽△ A1 B1C1 且 c = a1.∴ a1 b1 c1 (3)...