高二下數學空間平面問題 。

設平面E：Ax + By + Cz + D = 0

代入A(0,1,0) B(0,0,1)

B + D = 0, C + D = 0

=> B = C = -D

可令E：Ax + Dy + Dz - D = 0

E到C(1,2,3)的距離 |A + 2D + 3D - D|/√(A^2 + D^2+ D^2)

E到D(-2,-1,-4)的距離 |-2A - D - 4D - D|/√(A^2 + D^2+ D^2)

因此|A + 2D + 3D - D| = |-2A - D - 4D - D|

|A + 4D| = |-2A - 6D|

A + 4D = 2A + 6D 或 A + 4D = -2A - 6D

A = -2D 或 A = (-10/3)D

由此得E：10x - 3y -3z + 3 = 0 或 2x - y - z + 1 = 0