Question

37. Find the equation of the plane through the intersection of the planes x + 3y + 6 = 0 and
- Az = 0 and whose perpendicular distance from origin is unity​

Answer 1

Answer:

Any plane through the intersection of given planes is

(x+3y+6)+λ(3x−y−4z)=0

(1+3λ)x+(3−λ)y−4λz+6=0. ...(1)

Its perpendicular distance from (0,0,0) is 1.

∴

[(1+3λ)

2

+(3−λ)

2

+16λ

2

]

1/2

6

=1.

1+6λ+9λ

2

+9+λ

2

−6λ+16λ

2

=36

∴26λ

2

=26 or λ=±1.

Putting the value of λ in ( 1 ), the required equations are

2x+y−2z+3=0 and x−2y−2z−3=0.