The equation of the plane containing the lines , and parallel to the plane is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
The plane containing the line of intersection of 2x - 5y + z = 3 and x + y + 4z = 5 has the form:
(2x - 5y + z - 3) + λ(x + y + 4z - 5) = 0
This gives: (2+λ)x + (-5+λ)y + (1+4λ)z - (3+5λ) = 0
For this to be parallel to x + 3y + 6z = 1, the normal vectors must be proportional:
(2+λ)/1 = (-5+λ)/3 = (1+4λ)/6
From the first two ratios: 3(2+λ) = -5+λ → 6+3λ = -5+λ → 2λ = -11 → λ = -11/2
Check with first and third: 6(2+λ) = 1+4λ → 12+6λ = 1+4λ → 2λ = -11 → λ = -11/2 ✓
The constant term: 3 + 5λ = 3 + 5(-11/2) = 3 - 55/2 = -49/2
The normal direction becomes (2 - 11/2) = -7/2 for the x-coefficient. Multiplying through by -2/7 to normalize to coefficient 1 for x:
The plane is x + 3y + 6z = 49/7 = 7.
So the equation is x + 3y + 6z = 7.
(2x - 5y + z - 3) + λ(x + y + 4z - 5) = 0
This gives: (2+λ)x + (-5+λ)y + (1+4λ)z - (3+5λ) = 0
For this to be parallel to x + 3y + 6z = 1, the normal vectors must be proportional:
(2+λ)/1 = (-5+λ)/3 = (1+4λ)/6
From the first two ratios: 3(2+λ) = -5+λ → 6+3λ = -5+λ → 2λ = -11 → λ = -11/2
Check with first and third: 6(2+λ) = 1+4λ → 12+6λ = 1+4λ → 2λ = -11 → λ = -11/2 ✓
The constant term: 3 + 5λ = 3 + 5(-11/2) = 3 - 55/2 = -49/2
The normal direction becomes (2 - 11/2) = -7/2 for the x-coefficient. Multiplying through by -2/7 to normalize to coefficient 1 for x:
The plane is x + 3y + 6z = 49/7 = 7.
So the equation is x + 3y + 6z = 7.
[JEE Main 2015]