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JEE MainMathematicsThree Dimensional Geometry

Let L₁ be the line of intersection of the planes x + y - z = 0 and x + 2y = 5 . Let the point Q(1, , ) be the foot of the perpendicular drawn from the point P( , 4, 5) to the line L₁ . If L₂ is the line passing through the points P and Q , then the shortest distance between the line L₂ and the line L₃: x+1 1 = y-2 2 = z+3 3 is equal to:

Options

  1. A4 5
  2. B4 3
  3. C8 26 13
  4. D6 2

Correct answer

B. 4 3

Step-by-step solution

Since the point Q(1, , ) lies on the line L₁ , its coordinates must satisfy the equations of the planes: 1 + 2 = 5 2 = 4 = 2 1 + - = 0 1 + 2 - = 0 = 3 So, the point Q is (1, 2, 3) . The direction vector d ₁ of the line L₁ is parallel to the cross product of the normal vectors of the two planes: d ₁ = vmatrix i & j & k 1 & 1 & -1 1 & 2 & 0 vmatrix = i (0 - (-2)) - j (0 - (-1)) + k (2 - 1) = 2 i - j + k The vector PQ is given by (1 - ) i + (2 - 4) j + (3 - 5) k = (1 - ) i - 2 j - 2 k . Since Q is the foot of the perp

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