JEE Advanced2017MathematicsThree Dimensional GeometryActual
The equation of the plane passing through the point ( 1,1 , 1 ) and perpendicular to the planes 2 x + y - 2 z = 5 and 3 x - 6 y - 2 z = 7 is
Options
- A14 x + 2 y + 15 z = 31
- B14 x + 2 y - 15 z = 1
- C- 14 x + 2 y + 15 z = 3
- D14 x - 2 y + 15 z = 27
Correct answer
A. 14 x + 2 y + 15 z = 31
Step-by-step solution
The normal vector of required plane is parallel to vector = i ^ j ^ k ^ 2 1 - 2 3 - 6 -2 = - 14 i ^ - 2 j ^ - 15 k ^ ∴ The equation of required plane passing through (1, 1, 1) will be - 14 x - 1 - 2 y - 1 - 15 z - 1 = 0 ⇒ 14 x + 2 y + 15 z = 3 1 ∴ option 14 x + 2 y + 15 z = 31 is correct.