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 x−y+z=0. The equation of plane passing through the given planes is, ( x+y+z−1 )+λ( 2x+3y+4z−5 )=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 x−y+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 x−z+2=0 Therefore, the equation of plane is x−z+2=0.

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