Find the distance between the parallel planes
`vecr.(2hati-3hatj+6hatk) = 5` and
`vecr.(6hati-9hatj+18hatk) + 20 = 0`.

Options

A. `2/3`
B. `5/3`
C. `4/3`
D. `1/3`

Explanation

To find the distance between the parallel planes given by the equations:

1. \(\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 5\)
2. \(\vec{r} \cdot (6\hat{i} - 9\hat{j} + 18\hat{k}) + 20 = 0\)

we will follow these steps:

### Step 1: Convert the vector equations to Cartesian form

The first plane can be expressed in Cartesian form as follows:
\[ \vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 5 \implies 2x - 3y + 6z - 5 = 0 \]
This is the equation of Plane 1.

For the second plane:
\[ \vec{r} \cdot (6\hat{i} - 9\hat{j} + 18\hat{k}) + 20 = 0 \implies 6x - 9y + 18z + 20 = 0 \]
This is the equation of Plane 2.

### Step 2: Simplify the second plane's equation

Notice that the coefficients of the second plane can be simplified. We can factor out 3 from the second equation:
\[ 6x - 9y + 18z + 20 = 0 \implies 3(2x - 3y + 6z) + 20 = 0 \implies 2x - 3y + 6z + \frac{20}{3} = 0 \]
Thus, the equation of Plane 2 can be rewritten as:
\[ 2x - 3y + 6z + \frac{20}{3} = 0 \]

### Step 3: Identify coefficients for the distance formula

Now we have the two planes in the form:
1. \(2x - 3y + 6z - 5 = 0\) (Plane 1)
2. \(2x - 3y + 6z + \frac{20}{3} = 0\) (Plane 2)

Here, we can identify:
- \(d_1 = -5\)
- \(d_2 = \frac{20}{3}\)

### Step 4: Use the distance formula for parallel planes

The formula for the distance \(D\) between two parallel planes given by:
\[ ax + by + cz + d_1 = 0 \quad \text{and} \quad ax + by + cz + d_2 = 0 \]
is:
\[ D = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}} \]

Substituting the values:
- \(a = 2\), \(b = -3\), \(c = 6\)
- \(d_1 = -5\), \(d_2 = \frac{20}{3}\)

### Step 5: Calculate the distance

Calculate the numerator:
\[ |d_2 - d_1| = \left| \frac{20}{3} - (-5) \right| = \left| \frac{20}{3} + \frac{15}{3} \right| = \left| \frac{35}{3} \right| = \frac{35}{3} \]

Calculate the denominator:
\[ \sqrt{a^2 + b^2 + c^2} = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

Now, substituting these into the distance formula:
\[ D = \frac{\frac{35}{3}}{7} = \frac{35}{3 \times 7} = \frac{35}{21} = \frac{5}{3} \]

### Final Answer
The distance between the two parallel planes is \(\frac{5}{3}\) units.

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