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in Three Dimensional Geometry by (28.9k points)
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Let the equation of a plane be \(\bar r\). (2i – 3j + 5k) = 7, then

  1. Find the Cartesian equation of the plane. 
  2. Find the equation of a plane passing through the point (3, 4, -1) and parallel to the given plane. 
  3. Find the distance between the parallel planes.

1 Answer

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by (28.2k points)
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Best answer

1. Given, \(\bar r\).(2i – 3j + 5k) = 7 and if we substitute \(\bar r\) = xi + yj + zk Then we get the Cartesian equation as 2x – 3y + 5z – 7 = 0.

2. The equation of a plane parallel to the above plane differ only by a constant, therefore let the equation be 2x – 3y + 5z + k = 0.

⇒ 6 – 12 – 5 + k = 0 ⇒ k = 11

Therefore the equation is 2x – 3y + 5z + 11 = 0

3. The distance between the parallel planes

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