NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice
Consider the line L : x - 1 2 = y + 1 - 3 = z + 10 8 and a family of planes P containing the line L . The member of the family of planes P which is situated at a maximum distance from A 1 , 0 , 0 will be
Options
- Ax - 2 y - z = 13
- Bx + 2 y - z = 7
- C2 x + y - z = 7
- Dx + 2 y - 2 z = 13
Correct answer
A. x - 2 y - z = 13
Step-by-step solution
Let the foot of the perpendicular from A 1 , 0 , 0 to given line L = 0 is B , then the required plane must be perpendicular to A B and contain the given line. Any general point on the given line is B = 2 λ + 1 , - 3 λ - 1 , 8 λ - 10 A B → = < 2 λ , - 3 λ - 1 , 8 λ - 10 > ∴ A B → is perpendicular to the given line ⇒ 2 2 λ + - 3 - 3 λ - 1 + 8 8 λ - 10 = 0 ⇒ 4 λ + 9 λ + 3 + 64 λ - 80 = 0 ⇒ λ = 1 So, the coordinates of point B are 3 , - 4 , - 2 Equation of the required plane is 2 x - 1 - 4 y + 1 - 2 z + 10 = 0 ⇒ 2 x -