Find the vector and cartesian eq of the plane passing through the intersection of the planes r.(i+j+k)=6 and r.(2i+3j+4k)=-5 and the point (1,1,1)
Lets write these equations in cartesian form.
Plane 1 = x + y + z = 6
Plane 2 is 2x + 3y + 4z + 5 = 0
Plane passing through the intersection of these two planes is given by
x + y + z - 6 + λ( 2x + 3y + 4z + 5) = 0
or
( 1 + 2λ) x + ( 1 + 3λ)y + ( 1 +4λ) z +5λ - 6 = 0 ...............(1)
But it is given that this line pass trough (1,1,1). So (1,1,1) should satisfy in the above equation.
We get
1+2λ + 1 + 3λ + 1 + 4λ + 5λ - 6 = 0
or 14λ = 3
or λ = 3/14
Substituting the value of λ back in (1), we get
( 1 + 6/14) x + ( 1 + 9/14)y + ( 1+ 12/14) z + 15/14 - 6 = 0
or 20x + 23y + 26 z = 69 (you will get this on simplification).
The required vector equation is r.( 20i + 23j + 26k ) = 69
Hope that helps.
Plane 1 = x + y + z = 6
Plane 2 is 2x + 3y + 4z + 5 = 0
Plane passing through the intersection of these two planes is given by
x + y + z - 6 + λ( 2x + 3y + 4z + 5) = 0
or
( 1 + 2λ) x + ( 1 + 3λ)y + ( 1 +4λ) z +5λ - 6 = 0 ...............(1)
But it is given that this line pass trough (1,1,1). So (1,1,1) should satisfy in the above equation.
We get
1+2λ + 1 + 3λ + 1 + 4λ + 5λ - 6 = 0
or 14λ = 3
or λ = 3/14
Substituting the value of λ back in (1), we get
( 1 + 6/14) x + ( 1 + 9/14)y + ( 1+ 12/14) z + 15/14 - 6 = 0
or 20x + 23y + 26 z = 69 (you will get this on simplification).
The required vector equation is r.( 20i + 23j + 26k ) = 69
Hope that helps.