NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice
Distance between two non-intersecting planes P 1 and P 2 is 5 units, where P 1 is 2 x - 3 y + 6 z + 26 = 0 and P 2 is 4 x + b y + c z + d = 0 . The point A - 3 , 0 , - 1 lies between the planes P 1 and P 2 , then the value of 3 b + 4 c - 5 d is equal to
Options
- A580
- B120
- C- 18
- D- 120
Correct answer
B. 120
Step-by-step solution
Since both the planes are parallel P 1 : 4 x - 6 y + 12 z + 52 = 0 P 2 : 4 x - 6 y + 12 z + d = 0 Therefore, d - 52 14 = 5 ⇒ d - 52 = 70 ⇒ d = 122 , - 18 ∴ P 2 is 4 x - 6 y + 12 z + 122 = 0 or 4 x - 6 y + 12 z - 18 = 0 Since the point - 3,0 , - 1 is lying between P 1 and P 2 ∴ On substituting the point in both the equations of the plane, both the expressions must be of opposite signs Hence, 4 x - 6 y + 12 z - 18 = 0 is the equation of the required plane