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Question

Find the equation of the plane through the line of intersection of the planes and which is perpendicular to the plane

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Solution

It is given that the intersection of the planes x+y+z=1 and 2x+3y+4z=5 is perpendicular to the plane xy+z=0.

The equation of plane passing through the given planes is,

( x+y+z1 )+λ( 2x+3y+4z5 )=0 ( 2λ+1 )x+( 3λ+1 )y+( 4λ+1 )z( 5λ+1 )=0 (1)

Let a 1 , b 1 and c 1 be the direction ratios of the above plane. So,

a 1 =2λ+1 b 1 =3λ+1 c 1 =4λ+1

Let a 2 , b 2 and c 2 be the direction ratios of the plane xy+z=0. So,

a 2 =1 b 2 =1 c 2 =1

As the planes are perpendicular, then,

a 1 a 2 + b 1 b 2 + c 1 c 2 =0 ( 2λ+1 )( 1 )+( 3λ+1 )( 1 )+( 4λ+1 )( 1 )=0 3λ+1=0 λ= 1 3

Substitute λ= 1 3 in equation (1).

( 2( 1 3 )+1 )x+( 3( 1 3 )+1 )y+( 4( 1 3 )+1 )z( 5( 1 3 )+1 )=0 ( 2 3 +1 )x+( 1+1 )y+( 4 3 +1 )z( 5 3 +1 )=0 1 3 x 1 3 z+ 2 3 =0 xz+2=0

Therefore, the equation of plane is xz+2=0.


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