To find the equations for the planes parallel to the given plane and two units away from it, follow these steps:
Step 1: Write the given equation in general form.
The equation x + 2y - 2z = 1 can be rewritten as:
x + 2y - 2z - 1 = 0
Step 2: Find the normal vector of the given plane.
The coefficients of x, y, and z in the general form of the equation give us the normal vector. In this case, the normal vector is (1, 2, -2).
Step 3: Normalize the normal vector.
Divide the normal vector by its magnitude to obtain a unit vector. To normalize the vector (1, 2, -2), calculate its magnitude:
Magnitude = √(1^2 + 2^2 + (-2)^2) = √9 = 3
Dividing each component of the normal vector by its magnitude, we get the normalized normal vector: (1/3, 2/3, -2/3).
Step 4: Find the equation of the plane parallel to the given plane and two units away.
For a plane parallel to the given plane and displaced by distance d, the equation of the new plane is obtained by adding or subtracting d to the constant term of the general form of the equation.
Since the given plane is two units away, we will subtract 2 from the constant term (which is -1) to get -3.
Therefore, the equation for the plane parallel to x + 2y - 2z = 1 and two units away from it is:
x + 2y - 2z - 3 = 0
This is the equation of the desired plane.