Question

Find the equation of the plane passing through the points (2, -3, 1) and (-1, 1, -7) and perpendicular to the plane x – 2y + 5z + 1 = 0.

Answer 1

Required plane is perpendicular to the given plane x – 2y + 5z + 1 = 0
The required plane is parallel to the line which is perpendicular to the given plane.

Direction ratio of line a = 1, b = - 2, c = 5.
Hence, the required plane is

`|(x-x_1,y-y_1,z-z_1),(x_2-x_1, y_2-y_1, z_2-z_1),(a,b,c)|=0`

`|(x-2,y+ 3,z-1),(-1 -2, 1  + 3, -7 -1),(1,-2,5)|=0`

`|(x-2,y+ 3,z-1),(-3, 4, -8),(1,-2,5)|=0`

⇒ (x – 2) (20 – 16) – (y + 3) (-15 + 8) + (z – 1) (6 – 4 ) = 0
⇒ (x – 2) (4) – (y + 3) (-7) + (z – 1) (2) = 0
⇒ 4x + 7y + 2z + 11 = 0