用平面法向量解立体几何题

1、(江苏省苏州市迅达培训学校 ( 本讲适合高中数学竞赛试题中 , 几何题占有一定n BD , n DC1 3a x - 3ay =0, 3ax + a y + bz =量几何题量几何题y = 3x 3n = 3, 3, a 3b3要证平面外的直线 l 与平面, 只须出 l .例已知正三棱柱ABC - A1 B1 C1 中, 证明:建立如1的空间直角坐标系.nAB1= 3, 3, (0,a,b) =且AB1 BC1 DAB1 nAB1 BC1 若平面n1n2 则.A (0 ,0,0) 为n1 = ( x1 , y1 , z1) n2 = ( x2 , y2 , z2) . 由 n1 1 x + y

2、 =0, y + 1 z =A1 C1 , n1 A1 B , n2 AC, n2 AD1 ,- x1 + y1 = 0, y1 - z1 =0- x2 + y2 =0, - x2 + z2 =x1 = a , x2 = bn1 = ( a, a, a) , n2 = (b, b, b) y = - 1x = z 故 n = (2, - 1,2) A1 (1,0,1) P 0, 1 ,0 2A P - 1, 1 , - 1 易知 n1 = a n2 故 n1 n2 为共线向量. bn = - 2A1 Pn A1 P共线 1:, x1 = x2 = a ,n1 = n2 = ( a , a ,

3、a) . A1 BC1 与平面 D1 AC 有公共法向量 所以它们平行.1、12212a= 1,a = 1 222212121122n AC, n AB1 - x +2y =0,2y + z =x 2y 1z = - , n = (2,1, - 2) AB = (0,2,0) AB AB1 C | A1 B = (0,1, - 1) , A1 D1 = ( - 1,0,0) ABD1 A1 BD1 为n1 = ( x1 , y1 ,z1) , n2 = ( x2 , y2 , z2) 由 n1 AB , n1 AD1 , n2 A1 B , n2 A1 D1 1 arcsin | n| | 1

4、=0, - + =0 2=arcsin 1y2 - z2 =0, - x2 = 令x1 = a,则z1 = a3y2 = bz2 = 于是, n1 = ( a,0, a) , n2 = (0, b,b) ,设n1 8 -l 的 们的夹角与二面角的平 面角相等或互补. P n2 的夹角为.| n1| | = (a,0, a)(0,b, 2a2= 1 2, PA APB BAC l 于 C ,BC lACB , PA n1 , PB n2. 易知 APB ACB 互补.7 ABCD - A1 B1 C1 D1 二面A - BD1 - A1 的度数是 .(2004 , 高中数赛:设正方体为单位正方体

5、. 建立如9 的空间直角坐D(0,0,0) A (1 ,0,0) A - BD1 - A1 .A - D1 B - A1 60.:求二面角时 , 应注意法向量 n1 n2 的夹角是与其平面角相等还是,这可根据图62(2002 , 湖南省高中数学竞赛解:建立如10的空间直角坐标系 ,D(0,0,0) B1 (1,1,1) E 1, 1 21(b,0, c) , DB = (b, a,0) DC = (0, a ,0) A1 BD n = x , yz) nDA1 nDB得 bx + cz 0, bx+ ay 0.x = acy = - bcz = - ,1,0 2图所以, n = (ac , -

6、 bc , - ab)B1 E= 0, - 1 , - 1 2B F - 1 ,0, - 1 A1BDB1D1 CCA1 BDB1 EFn = xyzn B1 E, n B1 Fd=|nDC|=|(ac,- bc,- ab)(0,a,0)| a2b2a2b2 +b2c2+c2 - 1 y - z 0, - 1 x - z A1BDB1D1 Cz 1x = - 2, y = - , n = ( - 2, - 2,1又 DE = 1, 1 ,0 则点 D B1 2a2 ba2 b2 + b2 c2 + a2d=|nDE| | 求异面直线 若平面平面 , 求平面和平面之114A2 x + B2 y + C2 z + e =DP , , A