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JEE MainMathematicsThree Dimensional Geometry

Let the lines L₁ : x-1 1 = y+2 a = z+1 1 and L₂ : x-a 1 = y-3 -1 = z 1 intersect at a point P . The shortest distance of the point P from the plane 2x - 2y + z = 12 is:

Options

  1. A2
  2. B9
  3. C14
  4. D3

Correct answer

D. 3

Step-by-step solution

Any point on the first line L₁ is given by ( + 1, a - 2, - 1) . Any point on the second line L₂ is given by ( + a, - + 3, ) . Since the lines intersect, these points must coincide for some values of and : + 1 = + a a - 2 = - + 3 - 1 = Substituting = - 1 into the first equation: + 1 = ( - 1) + a a = 2 Substitute a = 2 and = - 1 into the second equation: 2 - 2 = -( - 1) + 3 2 - 2 = - + 4 3 = 6 = 2 The point of intersection P is obtained by substituting = 2 into the coordinates of L₁ : P(2 + 1, 2(2) - 2, 2 - 1) (3, 2,

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