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: Write the equations of the planes in standard form
The first plane is already in standard form:
\[ P_1: 2x + y + 2z - 8 = 0 \]

For the second plane, rearranging gives:
\[ P_2: 4x + 2y + 4z + 5 = 0 \implies 4x + 2y + 4z = -5 \]

### Step 2: Check if the planes are parallel
Two planes are parallel if their normal vectors are proportional. The normal vector for \(P_1\) is \((2, 1, 2)\) and for \(P_2\) is \((4, 2, 4)\).

We can see that:
\[ (4, 2, 4) = 2 \cdot (2, 1, 2) \]
Thus, the planes are indeed parallel.

### Step 3: Convert the equations to a common form
To find the distance between the two planes, we can express both planes in a similar form. We can multiply the first plane by 2:
\[ 2P_1: 4x + 2y + 4z = 16 \]

Now we have:
\[ P_1: 4x + 2y + 4z = 16 \]
\[ P_2: 4x + 2y + 4z = -5 \]

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

Here, \(D_1 = 16\) and \(D_2 = -5\). The coefficients \(A\), \(B\), and \(C\) are \(4\), \(2\), and \(4\) respectively.

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

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