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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

Lying in the plane x + y + z = 6 is a line L passing through 1,2 , 3 and perpendicular to the line of intersection of planes x + y + z = 6 and 2 x - y + z = 4 , then the equation of L is

Options

  1. Ax - 1 4 = y - 2 - 7 = z - 3 3
  2. Bx - 1 2 = y - 2 1 = z - 3 - 3
  3. Cx - 1 4 = y - 2 - 5 = z - 3 1
  4. Dx - 1 3 = y - 2 1 = z - 3 - 4

Correct answer

C. x - 1 4 = y - 2 - 5 = z - 3 1

Step-by-step solution

Required line lies in the plane x + y + z = 6 so it is perpendicular to i ^ + j ^ + k ^ and it is also perpendicular to line of intersection of the two given planes. Let, n → 1 = i ^ + j ^ + k ^ , n → 2 = 2 i ^ - j ^ + k ^ & ν → be a vector along the required line So, ν → = n → 1 × n → 1 × n → 2 = 2 i ^ + j ^ + k ^ - 3 2 i ^ - j ^ + k ^ = - 4 i ^ + 5 j ^ - k ^ Hence, equation of the line is x - 1 4 = y - 2 - 5 = z - 3 1

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