Question

Find the equation of the plane through the line of intersection of the planes 2x + 3y – z + 1 = 0 and x + y – 2z + 3 = 0 which is perpendicular to the plane 3x – y – 2z – 4 = 0 ?

Answer 1

We know that equation of plane passing through the line of intersection of planes

(a₁x + b₁y + c₁z + d₁) + k(a₂x + b₂y + c₂z + d₂) = 0

So equation of plane passing through the line of intersection of planes

2x + 3y – z + 1 = 0 and x + y – 2z + 3 = 0 is

(2x + 3y – z + 1) + k(x + y – 2z + 3) = 0

x(2 + k) + y(3 + k) + z( – 1 – 2k) + 1 + 3k = 0……(1)

Given that plane (1) is perpendicular if

a₁a₂ + b₁b₂ + c₁c₂ = 0 …… (2)

We know that two planes are perpendicular to plane,

3x – y – 2z – 4 = 0 …… (3)

Using (1) and (3) in eq. (2),

3(2 + k) + ( – 1)(3 + k) + ( – 2)( – 1 – 2k) = 0

6 + 3k – 3 – k + 2 + 4k = 0

6k + 5 = 0

6k = – 5

k = \(-\cfrac56\)

Put the value of k in equation (1),

x(2 + k) + y(3 + k) + z( – 1 – 2k) + 1 + 3k = 0

7 x + 13 y + 4 z – 9 = 0