The distance between the parallel planes 4x-2y+4+9=0 and 8x-4y+8z+21=0 is

Options

A. `1//4`
B. `1//2`
C. `3//2`
D. `7//4`

Answer

To find the distance between the two parallel planes given by the equations \(4x - 2y + 4z + 9 = 0\) and \(8x - 4y + 8z + 21 = 0\), we can follow these steps:

### Step 1: Identify the coefficients of the planes
The general form of a plane is given by \(Ax + By + Cz + D = 0\).
- For the first plane \(4x - 2y + 4z + 9 = 0\), we have:
- \(A_1 = 4\)
- \(B_1 = -2\)
- \(C_1 = 4\)
- \(D_1 = 9\)

- For the second plane \(8x - 4y + 8z + 21 = 0\), we can simplify it by dividing the entire equation by 2:
\[ 4x - 2y + 4z + \frac{21}{2} = 0 \]
Thus, we have:
- \(A_2 = 4\)
- \(B_2 = -2\)
- \(C_2 = 4\)
- \(D_2 = \frac{21}{2}\)

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

### Step 3: Substitute the values into the formula
Here, we have:
- \(D_1 = 9\)
- \(D_2 = \frac{21}{2}\)

Now, we calculate \(D_2 - D_1\):
\[ D_2 - D_1 = \frac{21}{2} - 9 = \frac{21}{2} - \frac{18}{2} = \frac{3}{2} \]

Next, we find the denominator \(\sqrt{A^2 + B^2 + C^2}\):
\[ A = 4, \quad B = -2, \quad C = 4 \]
\[ A^2 + B^2 + C^2 = 4^2 + (-2)^2 + 4^2 = 16 + 4 + 16 = 36 \]
\[ \sqrt{A^2 + B^2 + C^2} = \sqrt{36} = 6 \]

### Step 4: Calculate the distance
Now substituting back into the distance formula:
\[ d = \frac{\left|\frac{3}{2}\right|}{6} = \frac{3/2}{6} = \frac{3}{12} = \frac{1}{4} \]

### Conclusion
Thus, the distance between the two parallel planes is \(\frac{1}{4}\).