JEE MainMathematicsThree Dimensional Geometry
Let the line x-2 1 = y-3 1 = z-4 1 intersect the lines x-3 1 = y-2 -1 = z-6 2 and x-6 2 = y-6 1 = z-5 -1 at the points A and B respectively. If the distance of the mid-point of the line segment AB from the plane 2x - y + 2z + k = 0 is 3 , then the product of all possible values of k is
Options
- A-3
- B-72
- C135
- D63
Correct answer
D. 63
Step-by-step solution
Let the given line be L₁ : x-2 1 = y-3 1 = z-4 1 = . Any point on L₁ is ( +2, +3, +4) . For the intersection point A of L₁ and x-3 1 = y-2 -1 = z-6 2 = t , we equate the coordinates: +2 = t+3 - t = 1 +3 = -t+2 + t = -1 Solving these, we get = 0 and t = -1 . Substituting = 0 , the point A is (2, 3, 4) . For the intersection point B of L₁ and x-6 2 = y-6 1 = z-5 -1 = s , we equate the coordinates: +2 = 2s+6 - 2s = 4 +3 = s+6 - s = 3 Solving these, we get = 2 and s = -1 . Substituting = 2 , the point B is (4, 5, 6) .