JEE MainMathematicsThree Dimensional Geometry
Let the lines L₁ : x-1 2 = y-2 1 = z-3 -1 and L₂ : x-3 1 = y-3 2 = z-2 1 intersect at the point A . Let L be a line passing through A and parallel to the vector i - j + 2 k . If d is the shortest distance between the line L and the line L₃ : x-1 2 = y 1 = z-1 -1 , then the value of 35d^2 is
Options
- A256
- B400
- C490
- D25
Correct answer
A. 256
Step-by-step solution
Any point on L₁ is (2 +1, +2, - +3) and any point on L₂ is (t+3, 2t+3, t+2) . For the intersection point A , we equate the coordinates: 2 +1 = t+3 2 - t = 2 +2 = 2t+3 - 2t = 1 Solving these equations, we get = 1 and t = 0 . Substituting = 1 , the point A is (3, 3, 2) . The line L passes through A(3, 3, 2) and is parallel to b ₁ = i - j + 2 k . The line L₃ passes through P(1, 0, 1) and is parallel to b ₂ = 2 i + j - k . The shortest distance d between skew lines is given by: d = |( a ₂ - a ₁) ( b ₁ b ₂)| | b ₁ b ₂|