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JEE Main2014MathematicsThree Dimensional GeometryActual

Equation of the line of the shortest distance between the lines x ⁡ 1 = y ⁡ - 1 = z ⁡ 1 and x ⁡ - 1 0 = y ⁡ + 1 - 2 = z ⁡ 1 is

Options

  1. Ax ⁡ - 2 = y ⁡ 1 = z ⁡ 2
  2. Bx ⁡ 1 = y ⁡ - 1 = z ⁡ - 2
  3. Cx ⁡ - 1 1 = y ⁡ + 1 - 1 = z ⁡ - 2
  4. Dx ⁡ - 1 1 = y ⁡ + 1 - 1 = z ⁡ 1

Correct answer

C. x ⁡ - 1 1 = y ⁡ + 1 - 1 = z ⁡ - 2

Step-by-step solution

The equation of a line of the shortest distance between the given lines will be along the perpendicular to both the lines. A line perpendicular to L 1 :  x 1 = y - 1 = z 1 and L 2 :  x - 1 0 = y + 1 - 2 = z 1 is, i ^ j ^ k ^ 1 - 1 1 0 - 2 1 = i ^ - 1 + 2 - j ^ 1 - 0 + k ^ - 2 + 0 = i ^ − j ^ − 2 k ^ Let, x 1 = y - 1 = z 1 = α ⇒ x = α ,   y = - α ,   z = α Thus, a point on the line L 1 is P ( α ,   − α ,   α ) Similarly, let x -

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