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JEE Main202125 Feb 2021Morning ShiftMathematicsThree Dimensional GeometryActual

The equation of the line through the point 0 , 1 , 2 and perpendicular to the line x - 1 2 = y + 1 3 = z - 1 - 2 is :

Options

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

Correct answer

C. x - 3 = y - 1 4 = z - 2 3

Step-by-step solution

x - 1 2 = y + 1 3 = z - 1 - 2 = r ⇒   P x , y , z = 2 r + 1 , 3 r - 1 , - 2 r + 1 Since, Q P → ⊥ 2 i ^ + 3 j ^ - 2 k ^ ⇒   4 r + 2 + 9 r - 6 + 4 r + 2 = 0 ⇒   r = 2 17 ⇒   P 21 17 , - 11 17 , 13 17 ⇒   P Q → = 21 i ^ - 28 j ^ - 21 k ^ 17 So, Q P → : x - 3 = y - 1 4 = z - 2 3

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