Distance between two parallel planes `2x+y+2z=8 and 4x+2y+4z+5=0` is

Options

A. `(3)/(2)`
B. `(5)/(2)`
C. `(7)/(2)`
D. `(9)/(2)`

Answer

To find the distance between the two parallel planes given by the equations \(2x + y + 2z = 8\) and \(4x + 2y + 4z + 5 = 0\), we can follow these steps:

### Step 1: Rewrite the equations in standard form
The first plane is already in the standard form:
\[ P_1: 2x + y + 2z = 8 \]
For the second plane, we can rewrite it by moving the constant term to the other side:
\[ P_2: 4x + 2y + 4z = -5 \]

### Step 2: Make the coefficients of \(x\), \(y\), and \(z\) equal
To compare the two planes, we can multiply the first plane \(P_1\) by 2:
\[ 2(2x + y + 2z) = 2 \cdot 8 \implies 4x + 2y + 4z = 16 \]
Now we have:
\[ P_1: 4x + 2y + 4z = 16 \]
\[ P_2: 4x + 2y + 4z = -5 \]

### Step 3: Identify the coefficients
From both planes, we can identify:
- \(a = 4\)
- \(b = 2\)
- \(c = 4\)

### Step 4: Use the formula for the distance between two parallel planes
The distance \(d\) between two parallel planes of the form \(ax + by + cz = d_1\) and \(ax + by + cz = d_2\) is given by the formula:
\[ d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} \]
Here, \(d_1 = 16\) and \(d_2 = -5\).

### Step 5: Calculate the distance
Substituting the values into the formula:
\[ d = \frac{|16 - (-5)|}{\sqrt{4^2 + 2^2 + 4^2}} = \frac{|16 + 5|}{\sqrt{16 + 4 + 16}} = \frac{21}{\sqrt{36}} \]
Since \(\sqrt{36} = 6\), we have:
\[ d = \frac{21}{6} = \frac{7}{2} \]

### Final Answer
The distance between the two parallel planes is:
\[ \frac{7}{2} \]