Question

Find the equation of the plane passing through the line of intersection of planes 2x – y + z = 3 and 4x – 3y + 5z + 9 = 0 and parallel to the line `(x + 1)/2 = (y + 3)/4 = (z - 3)/5`

Answer 1

Given planes are

2x – y + z = 3 = 0,

4x – 3y + 5z + 9 = 0

Equation of required plane passing through their intersection is

(2x – y + z – 3) + λ(4x – 3y + 5z + 9) = 0 .....(1)

(2 + 4λ)x + (–1 – 3λ)y + (1 + 5λ)z + (–3 + 9λ) = 0

Direction ratios of the normal to the above plane are 2 + 4λ,  –1 – 3λ and 1 + 5λ

Plane is parallel to the line `(x + 1)/2 = (y + 3)/4 = (z - 3)/5`

Direction ratios of line are 2, 4, 5

Given that required plane is parallel to given line.

∴ Normal of the plane is perpendicular to the given line

2(2 + 4λ) + 4(–1 – 3λ) + 5(1 + 5λ) = 0

4 + 8λ – 4 – 12λ + 5 + 25λ = 0

21λ + 5 = 0

21λ = – 5

∴ λ = `-5/21`

Substituting λ in (1)

∴ Equation of plane is 

`(2x - y + z - 3)-5/21(4x - 3y + 5z + 9)` = 0

42x – 21y + 21z – 63 – 20x + 15y – 25z – 45 = 0

22x – 6y – 4z – 108 = 0

11x – 3y – 2z – 54 = 0