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NTA Abhyas JEE Main2020MathematicsThree Dimensional GeometryPractice

Let P 1 :   2 x + y + z + 1 = 0 , P 2 :   2 x - y + z + 3 = 0 and P 3 :   2 x + 3 y + z + 5 = 0 be three planes, then the distance of the line of intersection of planes P 1 = 0 and P 2 = 0 from the plane P 3 = 0 is

Options

  1. A3 14 units
  2. B6 14 units
  3. C3 7 units
  4. D6 7 units

Correct answer

B. 6 14 units

Step-by-step solution

A vector normal to P 1 = 0 is n 1 → = 2 i ^ + j ^ + k ^ A vector normal to P 2 = 0 is n 2 → = 2 i ^ - j ^ + k ^ Hence, the line of intersection is parallel to n 1 → × n 2 → = i ^ j ^ k ^ 2 1 1 2 - 1 1 = i ^ 2 - j ^ 0 + k ^ - 4 = 2 i ^ - 4 k ^ And for point of intersection Put z = 0 in P 1 = 0 and P 2 = 0 , i.e. 2 x + y + 1 = 0 2 x - y + 3 = 0 4 x + 4 = 0 ⇒ x = – 1 and y = 1 So point is - 1 , 1,0 Equation of the line of intersection is x + 1 1 = y - 1 0 = z - 0 - 2 The line is parallel to P 3 = 0 Hence, distance = -

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