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JEE Main202310 Apr 2023Evening ShiftMathematicsThree Dimensional GeometryActual

Let the line x 1 = 6 - y 2 = z + 8 5 intersect the lines x - 5 4 = y - 7 3 = z + 2 1 and x + 3 6 = 3 - y 3 = z - 6 1 at the points A and B respectively. Then the distance of the mid-point of the line segment A B from the plane 2 x - 2 y + z = 14 is

Options

  1. A3
  2. B11 3
  3. C4
  4. D10 3

Correct answer

C. 4

Step-by-step solution

Given, The line x 1 = 6 - y 2 = z + 8 5 intersect the lines x - 5 4 = y - 7 3 = z + 2 1 and x + 3 6 = 3 - y 3 = z - 6 1 at the points A and B respectively, Now plotting the diagram, we get, Now solving, x 1 = y - 6 - 2 = z + 8 5 = λ and x - 5 4 = y - 7 3 = z + 2 1 = t we get, A ≡ ( λ , - 2 λ + 6 , 5 λ - 8 ) ≡ ( 4 t + 5 , 3 t + 7 , t - 2 ) Now comparing both side and solving we get, t = - 1 , λ = 1 Hence, point   A ( 1 ,   4 ,   – 3 ) Now solving, x 1 = y - 6

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