The equation of the plane passing through the point and perpendicular to the planes and is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
The normal to the required plane is perpendicular to both given normals n₁ = (2,1,-2) and n₂ = (3,-6,-2), so it equals n₁ × n₂.
n₁ × n₂ = |i j k |
|2 1 -2|
|3 -6 -2|
= i(1·(-2) - (-2)·(-6)) - j(2·(-2) - (-2)·3) + k(2·(-6) - 1·3)
= i(-2 - 12) - j(-4 + 6) + k(-12 - 3)
= (-14, -2, -15)
So the normal direction is (14, 2, 15) (multiply by -1).
Plane through (1,1,1): 14(x-1) + 2(y-1) + 15(z-1) = 0 → 14x + 2y + 15z = 31.
n₁ × n₂ = |i j k |
|2 1 -2|
|3 -6 -2|
= i(1·(-2) - (-2)·(-6)) - j(2·(-2) - (-2)·3) + k(2·(-6) - 1·3)
= i(-2 - 12) - j(-4 + 6) + k(-12 - 3)
= (-14, -2, -15)
So the normal direction is (14, 2, 15) (multiply by -1).
Plane through (1,1,1): 14(x-1) + 2(y-1) + 15(z-1) = 0 → 14x + 2y + 15z = 31.
[JEE Advanced 2017]