Question

Find the equation of the plane passing through the line of intersection of the planes and and parallel to x -axis.

Answer 1

The given planes are :

r → ⋅( i ^ + j ^ + k ^ )=1 r → ⋅( i ^ + j ^ + k ^ )−1=0 (1)

r → ⋅( 2 i ^ +3 j ^ − k ^ )+4=0(2)

The equation of plane passing through the intersection of two planes ( A 1 x+ B 1 y+ C 1 z− d 1 =0 ) and ( A 2 x+ B 2 y+ C 2 z− d 2 =0 ) is given by,

( A 1 x+ B 1 y+ C 1 z− d 1 )+λ( A 2 x+ B 2 y+ C 2 z− d 2 )=0(3)

Substitute the values of equation (1) and equation (2) in equation (3).

r → ⋅( i ^ + j ^ + k ^ )−1+λ[ r → ⋅( 2 i ^ +3 j ^ − k ^ )+4 ]=0 r → ⋅[ ( 2λ+1 ) i ^ +( 3λ+1 ) j ^ +( 1−λ ) k ^ ]+( 4λ−1 )=0 (4)

The direction ratios of the new plane are given by,

( 2λ+1 ),( 3λ+1 ),( 1−λ )

The plane is parallel to the x axis. The normal of the plane is perpendicular to the x axis.

The direction ratios of x are 1, 0 and 0.

1⋅( 2λ+1 )+0⋅( 3λ+1 )+0⋅( 1−λ )=0 2λ+1=0 2λ=−1 λ=− 1 2 (5)

Substitute the value of λ from equation (5) to equation (4).

r → ⋅[ ( 2( −1 2 )+1 ) i ^ +( 2( −1 2 )+1 ) j ^ +( 1−( −1 2 ) ) k ^ ]+( 4( −1 2 )−1 )=0 r → ⋅[ ( −1+1 ) i ^ +( −3 2 +1 ) j ^ +( 1+ 1 2 ) k ^ ]+( −2−1 )=0 r → ⋅[ 0 i ^ − 1 2 j ^ + 3 2 k ^ ]−3=0 r → ⋅[ − j ^ +3 k ^ ]−6=0

Further simplify the above expression.

r → ⋅[ j ^ −3 k ^ ]+6=0

The above equation of plane in Cartesian form is given by,

y−3z+6=0

Thus, the equation of plane passing through the line of intersection of planes r → ⋅( i ^ + j ^ + k ^ )−1=0 and r → ⋅( 2 i ^ +3 j ^ − k ^ )+4=0 is y−3z+6=0.