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JEE MainMathematicsThree Dimensional Geometry

Let the lines L₁: x-1 2 = y+1 3 = z-2 1 and L₂: x-2 a = y-1 b = z-2 2 intersect at a point P . If the distance of the point P from the plane 2x + y - z = 5 is 6 , then the value of a + b is :

Options

  1. A-8
  2. B7
  3. C3
  4. D12

Correct answer

B. 7

Step-by-step solution

Let the point of intersection be P . Since P lies on L₁ , its coordinates can be written as (2 + 1, 3 - 1, + 2) . The perpendicular distance of P from the plane 2x + y - z - 5 = 0 is given as 6 . Using the distance formula: |2(2 + 1) + (3 - 1) - ( + 2) - 5| 2^2 + 1^2 + (-1)^2 = 6 |4 + 2 + 3 - 1 - - 2 - 5| 6 = 6 |6 - 6| = 6 | - 1| = 1 This gives - 1 = 1 = 2 or - 1 = -1 = 0 . If = 0 , the point P is (1, -1, 2) . Since P also lies on L₂ , substituting it into the equation of L₂ gives: 1 - 2 a = -1 - 1 b = 2 - 2 2 -1 a

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