The distance between the planes given by
`vecr.(hati+2hatj-2hatk)+5=0` and `vecr.(hati+2hatj-2hatk)-8=0` is

Options

A. 1 unit
B. `13/3` units
C. 13 units
D. none of these

Explanation

To find the distance between the two given planes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the equations of the planes**:
The equations of the planes are given as:
\[ \vec{r} \cdot (\hat{i} + 2\hat{j} - 2\hat{k}) + 5 = 0 \]
\[ \vec{r} \cdot (\hat{i} + 2\hat{j} - 2\hat{k}) - 8 = 0 \]

2. **Rewrite the equations in standard form**:
We can rewrite the equations in the form:
\[ \vec{r} \cdot \vec{n} = d_1 \quad \text{and} \quad \vec{r} \cdot \vec{n} = d_2 \]
where \(\vec{n} = \hat{i} + 2\hat{j} - 2\hat{k}\), \(d_1 = -5\), and \(d_2 = 8\).

3. **Check if the planes are parallel**:
The normal vectors of both planes are the same, \(\vec{n} = \hat{i} + 2\hat{j} - 2\hat{k}\). Since they are the same, the planes are parallel.

4. **Use the formula for the distance between two parallel planes**:
The distance \(D\) between two parallel planes can be calculated using the formula:
\[ D = \frac{|d_2 - d_1|}{|\vec{n}|} \]

5. **Calculate the values**:
- Calculate \(d_2 - d_1\):
\[ d_2 - d_1 = 8 - (-5) = 8 + 5 = 13 \]
- Calculate the magnitude of the normal vector \(|\vec{n}|\):
\[ |\vec{n}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

6. **Substitute the values into the distance formula**:
\[ D = \frac{|13|}{3} = \frac{13}{3} \]

### Final Answer:
The distance between the two planes is \(\frac{13}{3}\) units.

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