How can I determine the equation of a plane that passes through the point (4, 4, 5) and is parallel to the plane represented by the equation 3x + 4y + 3z = 5?

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To find the equation of a plane that passes through a given point and is parallel to another plane, we can follow these steps:

  1. Identify the Normal Vector: The equation of a plane in the general form is given as Ax + By + Cz = D, where (A, B, C) represents the normal vector of the plane. Given the plane equation 3x + 4y + 3z = 5, the normal vector is (3, 4, 3).
  2. Use the Point-Normal Form of the Plane Equation: The point-normal form of a plane’s equation can be expressed as A(x – x_0) + B(y – y_0) + C(z – z_0) = 0, where (x_0, y_0, z_0) is a point on the plane. In this case, our point is (4, 4, 5) and our normal vector is (3, 4, 3).
  3. Substitute the Values: Plugging these values into the point-normal form, we have: 
       3(x - 4) + 4(y - 4) + 3(z - 5) = 0
  4. Simplify the Equation: To simplify, expand the equation: 
       3x - 12 + 4y - 16 + 3z - 15 = 0

    This simplifies to: 

       3x + 4y + 3z - 43 = 0

    Alternatively, you can rewrite it as: 

       3x + 4y + 3z = 43

Therefore, the equation of the plane that passes through the point (4, 4, 5) and is parallel to the plane given by 3x + 4y + 3z = 5 is 3x + 4y + 3z = 43.

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