Question

Find the equation of the plane which contains the line of intersection of the planes , and which is perpendicular to the plane .

Answer 1

The equations of the given planes are,

r → ⋅( i ^ +2 j ^ +3 k ^ )−4=0(1)

r → ⋅( 2 i ^ + j ^ − k ^ )+5=0(2)

The equation of the plane perpendicular to the above planes are given by,

r → ⋅( 5 i ^ +3 j ^ −6 k ^ )+8=0(3)

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(4)

Substitute the equation of two planes from equation (1) and equation (2) in equation (4).

[ r → ⋅( i ^ +2 j ^ +3 k ^ )−4 ]+λ[ r → ⋅( 2 i ^ + j ^ − k ^ )+5 ]=0 r → ⋅[ ( 2λ+1 ) i ^ +( λ+2 ) j ^ +( 3−λ ) k ^ ]+( 5λ−4 )=0 (5)

The plane of equation (5) is perpendicular to the plane of equation (3).

5( 2λ+1 )+3( λ+2 )−6( 3−λ )=0 10λ+5+3λ+6−18+6λ=0 19λ−7=0 λ= 7 19

Substitute the value of λ in equation (5).

r → ⋅[ ( 2( 7 19 )+1 ) i ^ +( ( 7 19 )+2 ) j ^ +( 3−( 7 19 ) ) k ^ ]+( 5( 7 19 )−4 )=0 r → ⋅[ ( 14 19 +1 ) i ^ +( 7 19 +2 ) j ^ +( 3− 7 19 ) k ^ ]+( 35 19 −4 )=0 r → ⋅[ ( 33 19 ) i ^ +( 45 19 ) j ^ +( 50 19 ) k ^ ]− 41 19 =0 r → ⋅[ 33 i ^ +45 j ^ +50 k ^ ]−41=0

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

33x+45y+50z−41=0

Thus, the equation of plane which contains the line of intersection of planes r → ⋅( i ^ +2 j ^ +3 k ^ )−4=0 and r → ⋅( 2 i ^ + j ^ − k ^ )+5=0 and perpendicular to the plane r → ⋅( 5 i ^ +3 j ^ −6 k ^ )+8=0 is 33x+45y+50z−41=0.