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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

Consider the planes P 1 : 2 x - y + z = 6 and P 2 : x + 2 y - z = 4 having normals N → 1 and N → 2 respectively. The distance of the origin from the plane passing through the point 1,1 , 1 and whose normal is perpendicular to N 1 and N 2 is

Options

  1. A7 5 units
  2. B7 5 units
  3. C3 5 units
  4. D14 35 units

Correct answer

B. 7 5 units

Step-by-step solution

N → 1 = 2 i ^ - j ^ + k ^ (given) N → 2 = i ^ + 2 j ^ - k ^ (given) ∵ normal of required plane is N → 1 × N → 2 = i ^ j ^ k ^ 2 - 1 1 1 2 - 1 = i ^ - 1 - j ^ - 3 + k ^ 5 = - i ^ + 3 j ^ + 5 k ^ So, equation of the required plane is - 1 x - 1 + 3 y - 1 + 5 z - 1 = 0 - x + 3 y + 5 z - 7 = 0 x - 3 y - 5 z + 7 = 0 So, distance from 0,0 , 0 is = 0 - 0 + 0 + 7 35 = 7 5

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