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11,当且仅当a?b?c?时取“=”. ………………5分
3271112222(2)柯西不等式a?b?c?(a?b?c)?,由(1)知3abc?
333?abc??a2?b2?c2?3abc,当且仅当a?b?c时取“=”. ………………10分
?y?2x?4m2n222.解:(1)?2?A(4,4),B(1,?2),设A1(,m),B1(,n),
44y?4x?kAT?kA1T?4m?2?m2?4t?4m?tm?m(m?4)?t(4?m) 4?tm?t4t2?m??t?A1(,?t),同理:B1(t2,2t)?k?43tt2t?42?4?kt?4定值.…5分 t4t22(2)A1B1:y?2t?(x?t),令y?0得N(,0),而M(2,0)
t2t2?ttt2S4TN?yA1t2S1yA12????S4?S1 ??2?S2?S1,?S1TM?yAt?2488S2yB2t(t?2)2tt2S3TN?yB1t22?????S3?S1 S1TM?yAt?2444S1,S2,S3,S4构成的等比数列,∴t2?1而t?0?t?1. ………………10分
23.解:如图以CB、CA分别为x,y轴,过C作直线Cz//BC1,以Cz为z轴
?B(3,0,0),C(0,0,0),A(0,3,0),C1(3,0,3)
B1zA1C1CB1?CC1?CB?(6,0,3)?B1(6,0,3) CA1?CC1?CA?(3,3,3)?A1(3,3,3)
(1)T是△ABC1重心?T(2,1,1)?TA1,2,2) 1?(设面ABC1的法向量为n1?(x1,y1,z1),AB?(3,?3,0)
yAxTBC
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?3x1?3y1?0?z1?0?????取法向量n1?(1,1,0) ?3x1?3y1?3z1?0?x1?y1?cos?TA1,n1??33?2?2???TA1,n1?? 24设TA1与面ABC1所成角为?????2??TA1,n1???4. ………………5分
(2)T在面ABC1内,CT?CB?BT?CB?mBC1?nBA??3?3n,3n,3m?,
即T(3?3n,3n,3m).由TB1?TC得
(3?3n)2?(3n)2?(3m)2?(3n?3)2?(3n)2?(3m?3)2??2m?4n??1①
设面CAA1C1法向量为n2?(x2,y2,z2),CA?(0,3,0),CC1?(3,0,3)
?3y2?0???取n2?(1,0,?1) ?3x2?3z2?0设面TA1C1法向量为n3?(x3,y3,z3),C1A1?(0,3,0),CT1?(?3n,3n,3m?3)
?y3?0由平面TAC???取n3?(m?1,0,n),11?平面ACC1A1得
??3nx3?(3m?3)z3?0cos?n2,n3??m?1?n2?(m?1)?n22?0?m?n?1②
由①②解得n?13311?339?,m?,?存在点T?,,?,TC=. ………10分 222222??
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