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))` unit

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: Write the equations in standard form
The standard form of a plane is given by \(Ax + By + Cz + D = 0\).

1. For the first plane \(2x + 3y + 4z = 4\):
\[ 2x + 3y + 4z - 4 = 0 \]
Here, \(A_1 = 2\), \(B_1 = 3\), \(C_1 = 4\), and \(D_1 = -4\).

2. For the second plane \(4x + 6y + 8z = 12\):
\[ 4x + 6y + 8z - 12 = 0 \]
We can simplify this by dividing the entire equation by 2:
\[ 2x + 3y + 4z - 6 = 0 \]
Here, \(A_2 = 2\), \(B_2 = 3\), \(C_2 = 4\), and \(D_2 = -6\).

### Step 2: Check if the planes are parallel
The normal vectors of the planes can be derived from the coefficients of \(x\), \(y\), and \(z\):
- Normal vector of the first plane: \(\vec{n_1} = (2, 3, 4)\)
- Normal vector of the second plane: \(\vec{n_2} = (2, 3, 4)\)

Since \(\vec{n_1}\) and \(\vec{n_2}\) are proportional, the planes are parallel.

### Step 3: Calculate the distance between the two parallel planes
The formula for the distance \(d\) between two parallel planes \(Ax + By + Cz + D_1 = 0\) and \(Ax + By + Cz + D_2 = 0\) is given by:
\[ d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}} \]

Substituting the values we found:
- \(D_1 = -4\)
- \(D_2 = -6\)
- \(A = 2\), \(B = 3\), \(C = 4\)

### Step 4: Substitute into the distance formula
\[ d = \frac{|-4 - (-6)|}{\sqrt{2^2 + 3^2 + 4^2}} \]
\[ = \frac{|-4 + 6|}{\sqrt{4 + 9 + 16}} \]
\[ = \frac{|2|}{\sqrt{29}} \]
\[ = \frac{2}{\sqrt{29}} \]

### Final Answer
Thus, the distance between the two planes is:
\[ \frac{2}{\sqrt{29}} \text{ units} \]

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