Question:
The equation of the plane containing the line 2x-5y z=3 , x y 4z=5 and parallel to the plane x 3y 6z=1 is
Answer:
Let the equation of the plane is
(2x - 5y + z - 3) + λ(x + y + 4z - 5) = 0
=> (2 + λ)x + (λ - 5)y + (4λ + 1)z - (3 + 5λ) = 0
Since the plane is parallel to x + 3y + 6z - 1 = 0
=> (2 + λ)/1 = (λ - 5)/3 = (1 + 4λ)/6
=> 6 + 3λ = λ - 5
=> 2λ = -11
=> λ = -11/2
Again,
6λ - 30 = 3 + 12λ
=> -6λ = -33
=> λ = -33/6
=> λ = -11/2
So, the required equation of plane is
(2x - 5y + z - 3) + (-11/2)*(x + y + 4z - 5) = 0
=> 2(2x - 5y + z - 3) + (-11)*(x + y + 4z - 5) = 0
=> 4x - 10y + 2z - 6 - 11x - 11y - 44z + 55 = 0
=> -7x - 21y - 42z + 49 = 0
=> x + 3y + 6z - 7 = 0
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