What is the distance between the planes `x-2y+z-1=0 and -3x+6y-3z+2=0`?

Options

A. 3 units
B. 1 unit
C. 0
D. None of these

Answer

To find the distance between the two planes given by the equations \( x - 2y + z - 1 = 0 \) and \( -3x + 6y - 3z + 2 = 0 \), we can follow these steps:

### Step 1: Rewrite the equations of the planes
First, we can rewrite the equations of the planes in the standard form \( ax + by + cz = d \).

1. For the first plane:
\[ x - 2y + z = 1 \]
This can be written as:
\[ 1x - 2y + 1z = 1 \]
Here, \( a_1 = 1, b_1 = -2, c_1 = 1, d_1 = 1 \).

2. For the second plane:
\[ -3x + 6y - 3z = -2 \]
This can be written as:
\[ -3x + 6y - 3z = -2 \]
Here, \( a_2 = -3, b_2 = 6, c_2 = -3, d_2 = -2 \).

### Step 2: Check if the planes are parallel
To determine if the planes are parallel, we compare the coefficients of \( x, y, z \).

The normal vector of the first plane is \( (1, -2, 1) \) and for the second plane is \( (-3, 6, -3) \).

We can see that the second plane's normal vector is a scalar multiple of the first plane's normal vector:
\[ (-3, 6, -3) = -3 \cdot (1, -2, 1) \]
Thus, the planes are parallel.

### Step 3: Use the distance formula for parallel planes
The distance \( D \) between two parallel planes given by the equations \( ax + by + cz = d_1 \) and \( ax + by + cz = d_2 \) is given by:
\[ D = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \]

### Step 4: Substitute the values
From our equations, we have:
- \( d_1 = 1 \)
- \( d_2 = -2 \)
- Coefficients \( a = 1, b = -2, c = 1 \)

Now, substituting these values into the distance formula:
\[ D = \frac{|-2 - 1|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|-3|}{\sqrt{1 + 4 + 1}} = \frac{3}{\sqrt{6}} \]

### Step 5: Simplify the result
To simplify \( \frac{3}{\sqrt{6}} \):
\[ D = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2} \]

### Final Answer
Thus, the distance between the two planes is:
\[ \frac{\sqrt{6}}{2} \]