Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector `3 hat i+5 hat j-6 hat k`.

Options

A. `6sqrt70`
B. `5sqrt70`
C. `8sqrt70`
D. `7sqrt70`

Explanation

To find the vector equation of a plane that is at a distance of 7 units from the origin and normal to the vector \( \mathbf{n} = 3\hat{i} + 5\hat{j} - 6\hat{k} \), we can follow these steps:

### Step 1: Identify the normal vector and its magnitude
The normal vector to the plane is given as \( \mathbf{n} = 3\hat{i} + 5\hat{j} - 6\hat{k} \).

To find the magnitude of the normal vector \( \mathbf{n} \):
\[ |\mathbf{n}| = \sqrt{3^2 + 5^2 + (-6)^2} = \sqrt{9 + 25 + 36} = \sqrt{70} \]

### Step 2: Find the unit normal vector
The unit normal vector \( \hat{n} \) is obtained by dividing the normal vector by its magnitude:
\[ \hat{n} = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}} \]

### Step 3: Use the formula for the distance from a point to a plane
The distance \( D \) from the origin to the plane can be expressed as:
\[ D = \frac{|\mathbf{r} \cdot \hat{n}|}{|\hat{n}|} \]
Given that \( D = 7 \), we can set up the equation:
\[ 7 = \frac{|\mathbf{r} \cdot \hat{n}|}{1} \quad \text{(since } |\hat{n}| = 1\text{)} \]
This simplifies to:
\[ |\mathbf{r} \cdot \hat{n}| = 7 \]

### Step 4: Substitute the unit normal vector into the equation
Now substituting \( \hat{n} \):
\[ |\mathbf{r} \cdot \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}}| = 7 \]
Multiplying both sides by \( \sqrt{70} \):
\[ |\mathbf{r} \cdot (3\hat{i} + 5\hat{j} - 6\hat{k})| = 7\sqrt{70} \]

### Step 5: Write the vector equation of the plane
The vector equation of the plane can be expressed as:
\[ \mathbf{r} \cdot (3\hat{i} + 5\hat{j} - 6\hat{k}) = 7\sqrt{70} \]

### Final Answer
Thus, the vector equation of the plane is:
\[ \mathbf{r} \cdot (3\hat{i} + 5\hat{j} - 6\hat{k}) = 7\sqrt{70} \]