说明平面通过坐标原点；(1) D = 0, Ax + By + Cz = 0,(2) A = 0, By +Cz + D = 0, : n = (0,B,C) I (1,0,0),D±0,By+Cz+D=0,平面平行于x轴：D =0,Bv+Cz =0,平面过x 轴;(3) A = B = 0,Cz + D = 0,: n= (0,0,C) 1 (1,0,0),n=(0,0,C) (0,1,0),D±0,Cz+D=0,平面平行于xy面；D=0,z=0,平面与 xy面重合其他情形类推o08个不不高等教学教学部不不
高等数学教学部 6 (1) D 0, Ax By Cz 0, (2) A 0,By Cz D 0, n (0,B,C) (1,0,0), (3) A B 0,Cz D 0, n (0,0,C) (1,0,0),n (0,0,C) (0,1,0),
S例 1 求过三点 A(2,-1,4)、B(-1,3,-2)和C(0,2,3)的平面方程解AC = (-2, 3,-1)AB = (-3, 4,-6)C取 n = AB × AC = (14 , 9,-1),14(x - 2) + 9(y +1) - (z - 4) = 0, 14x + 9y - z -15 = 0解二设所求平面方程为Ax+By+Cz+D=0.因平面过三点A(2,-1,4)、B(-1,3,-2)和C(0,2,3)，得[A= -14C2A-B+4C+D=0-A+3B-2C+D=0 =3 B=-9C.:. 14x + 9 y - z - 15 = 02B+3C+D=0D =15C001018个不不高教学教学部不不不
高等数学教学部 7 AB ( 3 , 4 , 6 ) AC ( 2 , 3 , 1 ) 取 n AB AC (14 , 9 , 1 ), 14(x 2) 9( y 1) (z 4) 0, 14x 9 y z 15 0. Ax By Cz D 0, 2 3 0 3 2 0 2 4 0 B C D A B C D A B C D D C B C A C 15 9 14 14x 9 y z 15 0