Solveeit Logo

Question

Question: The equation of the plane passing through the intersection of the planes \(x + 2y + 3z + 4 = 0\) and...

The equation of the plane passing through the intersection of the planes x+2y+3z+4=0x + 2y + 3z + 4 = 0 and 4x+3y+2z+1=04x + 3y + 2z + 1 = 0 and the origin is

A

3x+2y+z+1=03x + 2y + z + 1 = 0

B

3x+2y+z=03x + 2y + z = 0

C

2x+3y+z=02x + 3y + z = 0

D

x+y+z=0x + y + z = 0

Answer

3x+2y+z=03x + 2y + z = 0

Explanation

Solution

Equation of plane passing through the intersection of given planes, is

(x+2y+3z+4)+λ(4x+3y+2z+1)=0( x + 2 y + 3 z + 4 ) + \lambda ( 4 x + 3 y + 2 z + 1 ) = 0 .....(i)

Plane (i) is passing through origin i.e., (0, 0, 0)

4+λ=04 + \lambda = 0λ=4\lambda = - 4

Put the value of λ\lambda in (i),

15x10y5z=0- 15 x - 10 y - 5 z = 03x+2y+z=03 x + 2 y + z = 0