The distance between the planes `2x+2y-z+2=0 and 4x+4y-2z+5=0` is

Options

A. `3/2` units
B. `1/2` unit
C. `1/6` unit
D. none of these

Answer

To find the distance between the two planes given by the equations \(2x + 2y - z + 2 = 0\) and \(4x + 4y - 2z + 5 = 0\), we can follow these steps:

### Step 1: Identify the equations of the planes
The equations of the planes are:
1. Plane 1: \(2x + 2y - z + 2 = 0\)
2. Plane 2: \(4x + 4y - 2z + 5 = 0\)

### Step 2: Check if the planes are parallel
To check if the planes are parallel, we can compare the coefficients of \(x\), \(y\), and \(z\) in both equations.

For Plane 1, the coefficients are:
- \(a_1 = 2\)
- \(b_1 = 2\)
- \(c_1 = -1\)

For Plane 2, the coefficients are:
- \(a_2 = 4\)
- \(b_2 = 4\)
- \(c_2 = -2\)

We can see that Plane 2 can be obtained by multiplying the coefficients of Plane 1 by 2. Thus, the planes are parallel.

### Step 3: Rewrite Plane 2 in a comparable form
We can rewrite Plane 2 to have the same coefficients as Plane 1. Multiply the first plane equation by 2:
\[ 2(2x + 2y - z + 2) = 4x + 4y - 2z + 4 = 0 \]
Now we have:
1. Plane 1: \(2x + 2y - z + 2 = 0\)
2. Plane 2: \(4x + 4y - 2z + 4 = 0\)

### Step 4: Use the formula for the distance between two parallel planes
The formula for the distance \(d\) between two parallel planes of the form \(Ax + By + Cz + D_1 = 0\) and \(Ax + By + Cz + D_2 = 0\) is given by:
\[ d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]

### Step 5: Identify \(D_1\) and \(D_2\)
From the equations:
- For Plane 1, \(D_1 = 2\)
- For Plane 2, \(D_2 = 4\)

### Step 6: Substitute values into the distance formula
Substituting the values into the formula:
\[ d = \frac{|4 - 2|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{2}{\sqrt{4 + 4 + 1}} = \frac{2}{\sqrt{9}} = \frac{2}{3} \]

### Step 7: Final answer
Thus, the distance between the two planes is:
\[ \frac{2}{3} \]