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

Options

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

Answer

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

### Step 1: Identify the equations of the planes
The given planes are:
1. \(2x + 3y + 4z = 4\) (Plane 1)
2. \(4x + 6y + 8z = 12\) (Plane 2)

### Step 2: Simplify the second plane
Notice that the second plane can be simplified. We can divide the entire equation of Plane 2 by 2:
\[ 4x + 6y + 8z = 12 \implies 2x + 3y + 4z = 6 \]
Now we have:
1. \(2x + 3y + 4z = 4\) (Plane 1)
2. \(2x + 3y + 4z = 6\) (Plane 2)

### Step 3: Identify the coefficients
From the equations, we can identify:
- \(a = 2\)
- \(b = 3\)
- \(c = 4\)
- \(d_1 = -4\) (from Plane 1)
- \(d_2 = -6\) (from Plane 2)

### 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: Calculate \(d_2 - d_1\)
Calculate the absolute difference:
\[ |d_2 - d_1| = |-6 - (-4)| = |-6 + 4| = |-2| = 2 \]

### Step 6: Calculate \(a^2 + b^2 + c^2\)
Now calculate \(a^2 + b^2 + c^2\):
\[ a^2 + b^2 + c^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29 \]

### Step 7: Substitute into the distance formula
Now substitute the values into the distance formula:
\[ D = \frac{2}{\sqrt{29}} \]

### Step 8: Final answer
Thus, the distance between the two planes is:
\[ D = \frac{2}{\sqrt{29}} \]