Distance between two planes :
2x + 3y + 4z = 5 and 4x + 6y + 8z = 12 is :

Options

A. 2 units
B. 4 units
C. 8 units
D. `(1)/(29)` units.

Answer

To find the distance between the two planes given by the equations \(2x + 3y + 4z = 5\) and \(4x + 6y + 8z = 12\), we can follow these steps:

### Step 1: Identify the equations of the planes
The first plane is given by:
\[ P_1: 2x + 3y + 4z = 5 \]
The second plane is given by:
\[ P_2: 4x + 6y + 8z = 12 \]

### Step 2: Simplify the second plane
We can simplify the second plane \(P_2\) by dividing all terms by 2:
\[ P_2: 2x + 3y + 4z = 6 \]

### Step 3: Compare the two planes
Now we have:
\[ P_1: 2x + 3y + 4z = 5 \]
\[ P_2: 2x + 3y + 4z = 6 \]
Both planes have the same normal vector \((2, 3, 4)\), which indicates that they are parallel.

### Step 4: Calculate the distance between the two parallel planes
The formula to find the distance \(d\) between two parallel planes of the form \(Ax + By + Cz = D_1\) and \(Ax + By + Cz = D_2\) is given by:
\[ d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]
In our case:
- \(D_1 = 5\)
- \(D_2 = 6\)
- \(A = 2\), \(B = 3\), \(C = 4\)

### Step 5: Substitute values into the formula
Now substituting the values into the formula:
\[ d = \frac{|6 - 5|}{\sqrt{2^2 + 3^2 + 4^2}} \]
Calculating the numerator:
\[ |6 - 5| = 1 \]
Calculating the denominator:
\[ \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

### Step 6: Final calculation of distance
Thus, the distance \(d\) is:
\[ d = \frac{1}{\sqrt{29}} \]

### Conclusion
The distance between the two planes is:
\[ \frac{1}{\sqrt{29}} \text{ units} \]