The distance between the planes `x+2y-2z+1=0` and `2x+4y-4z+5=0`, is

Options

A. 4 units
B. 2 units
C. `1/2` units
D. `1/4` units

Explanation

To find the distance between the two given planes, we first need to confirm that they are parallel. The equations of the planes are:

1. Plane 1: \( x + 2y - 2z + 1 = 0 \)
2. Plane 2: \( 2x + 4y - 4z + 5 = 0 \)

### Step 1: Check if the planes are parallel

Two planes are parallel if their normal vectors are proportional. The normal vector of a plane given by the equation \( Ax + By + Cz + D = 0 \) is \( (A, B, C) \).

- For Plane 1, the normal vector is \( (1, 2, -2) \).
- For Plane 2, the normal vector is \( (2, 4, -4) \).

To check if they are proportional, we can see if there exists a constant \( k \) such that:

\[ (2, 4, -4) = k(1, 2, -2) \]

By inspection, we can see that \( k = 2 \) works:

\[ 2 = 2 \cdot 1, \quad 4 = 2 \cdot 2, \quad -4 = 2 \cdot -2 \]

Since the normal vectors are proportional, the planes are indeed parallel.

### Step 2: Find the distance between the 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:

\[ d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}} \]

Here, we need to rewrite both planes in the standard form:

- Plane 1: \( x + 2y - 2z + 1 = 0 \) implies \( D_1 = 1 \).
- Plane 2: \( 2x + 4y - 4z + 5 = 0 \) can be rewritten as \( x + 2y - 2z + \frac{5}{2} = 0 \) (dividing the entire equation by 2), so \( D_2 = \frac{5}{2} \).

### Step 3: Apply the distance formula

Now, we can substitute \( A = 1 \), \( B = 2 \), \( C = -2 \), \( D_1 = 1 \), and \( D_2 = \frac{5}{2} \) into the distance formula:

\[ d = \frac{\left| \frac{5}{2} - 1 \right|}{\sqrt{1^2 + 2^2 + (-2)^2}} \]

Calculating the numerator:

\[ \frac{5}{2} - 1 = \frac{5}{2} - \frac{2}{2} = \frac{3}{2} \]

Calculating the denominator:

\[ \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

Now substituting back into the distance formula:

\[ d = \frac{\left| \frac{3}{2} \right|}{3} = \frac{3}{2 \cdot 3} = \frac{3}{6} = \frac{1}{2} \]

### Final Answer

The distance between the planes is \( \frac{1}{2} \).

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