The vector equation of the plane through the line of intersection of the planes and , which is perpendicular to the plane is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
The family of planes through the line of intersection is (x+y+z-1) + λ(2x+3y+4z-5) = 0, giving normal vector n = (1+2λ)i + (1+3λ)j + (1+4λ)k.
For perpendicularity with x-y+z=0 (normal: i-j+k), dot product = 0:
(1+2λ)(1) + (1+3λ)(-1) + (1+4λ)(1) = 0
1+2λ - 1-3λ + 1+4λ = 0
1 + 3λ = 0 → λ = -1/3
Substituting: (1-2/3)x + (1-1)y + (1-4/3)z = 1-5/3
(1/3)x + 0·y + (-1/3)z = -2/3
Multiply by 3: x - z = -2, i.e., x - z + 2 = 0.
In vector form: r·(i - k) + 2 = 0.
For perpendicularity with x-y+z=0 (normal: i-j+k), dot product = 0:
(1+2λ)(1) + (1+3λ)(-1) + (1+4λ)(1) = 0
1+2λ - 1-3λ + 1+4λ = 0
1 + 3λ = 0 → λ = -1/3
Substituting: (1-2/3)x + (1-1)y + (1-4/3)z = 1-5/3
(1/3)x + 0·y + (-1/3)z = -2/3
Multiply by 3: x - z = -2, i.e., x - z + 2 = 0.
In vector form: r·(i - k) + 2 = 0.
[JEE Main 2019 · 8 Apr · Shift II]