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Let P 1 : x - 2 y + 3 z = 5 and P 2 : 2 x - 3 y + z + 4 = 0 be two planes. The equation of the plane perpendicular to the line of intersection of P 1 = 0 and P 2 = 0 and passing through 1,1 , 1 is

Options

  1. A11 x - 5 y + 7 z - 13 = 0
  2. B7 x + 5 y + z =13
  3. Cx + 2 y + z - 4 = 0
  4. Dx - 2 y + 4 z + 3 = 0

Correct answer

B. 7 x + 5 y + z =13

Step-by-step solution

The normal vector of plane P 1 = 0 is n 1 → = i ^ - 2 j ^ + 3 k ^ The normal vector of plane P 2 = 0 is n 2 → = 2 i ^ - 3 j ^ + k ^ A vector parallel to the line of intersection of P 1 = 0 and P 2 = 0 is i ^ j ^ k ^ 1 - 2 3 2 - 3 1 = 7 i ^ + 5 j ^ + k ^ It is also a normal vector for the required plane Hence, the equation of the required plane is 7 x - 1 + 5 y - 1 + z - 1 = 0 7 x + 5 y + z - 13 = 0

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