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KCET2024MathematicsThree Dimensional Geometry

The distance between the two planes 2 x+3 y+4 z=4 and 4 x+6 y+8 z=12 is

Options

  1. A2 units
  2. B8 units
  3. C2 29 unit
  4. D4 units

Correct answer

C. 2 29 unit

Step-by-step solution

The equations of planes are 2 x+3 y+4 z=4 ....(i) 4 x+6 y+8 z=122 x+3 y+4 z=6 ....(ii) It can be seen that the given planes are parallel. It is known that the distance between two parallel planes is | d₂-d₁ a^2+b^2+c^2 | . So, D= | 6-4 (2)^2+(3)^2+(4)^2 |= 2 29 units

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