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JEE Main20206 Sep 2020Morning ShiftMathematicsThree Dimensional GeometryActual

The shortest distance between the lines x - 1 0 = y + 1 - 1 = z 1 and x + y + z + 1 = 0 , 2 x - y + z + 3 = 0 is

Options

  1. A1
  2. B1 3
  3. C1 2
  4. D1 2

Correct answer

B. 1 3

Step-by-step solution

We know that shortest distance between two skew lines exists along the line which is perpendicular to both the lines. Now finding the equation of a plane P , whose normal is perpendicular to the given first line and the line obtained by planes x + y + z + 1 = 0 , 2 x - y + z + 3 = 0 . So, P is: x + y + z + 1 + λ ( 2 x - y + z + 3 ) = 0 It should be parallel to given line. ⇒   0 ( 1 + 2 λ ) - 1 ( 1 - λ ) + 1 ( 1 + λ ) = 0 ⇒ λ = 0 Thus, Plane, P   is   x + y + z + 1

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