NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice
The distance of the point 2,3 , 2 from the plane 3 x + 4 y + 4 z = 23 measured parallel to the line x + 3 1 = y - 6 - 2 = z - 1 1 is
Options
- A108 units
- B12 units
- C54 units
- D236 units
Correct answer
C. 54 units
Step-by-step solution
Equation of the line passing through 2,3 , 2 and parallel to the given line is x - 2 1 = y - 3 - 2 = z - 2 1 Any general point on this line is λ + 2 , - 2 λ + 3 , λ + 2 This must satisfy the given equation of the plane ⇒ 3 λ + 6 - 8 λ + 12 + 4 λ + 8 = 23 ⇒ - λ + 26 = 23 ⇒ λ = 3 The point on the plane is 5 , - 3 ,5 Hence, the required distance = 9 + 36 + 9 = 54 units