Find the vector equation of a plane which is at a distance of 5 units from the origin and whose normal vector is ` 2 hati - hatj + 2 hatk`.

Answer

To find the vector equation of a plane that is at a distance of 5 units from the origin and has a normal vector of \( \vec{n} = 2 \hat{i} - \hat{j} + 2 \hat{k} \), we can follow these steps:

### Step 1: Identify the normal vector and its components
The normal vector \( \vec{n} \) is given as:
\[ \vec{n} = 2 \hat{i} - \hat{j} + 2 \hat{k} \]
This means that the components of the normal vector are \( A = 2 \), \( B = -1 \), and \( C = 2 \).

### Step 2: Use the formula for the distance from a point to a plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane defined by the equation \( Ax + By + Cz - D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 - D|}{\sqrt{A^2 + B^2 + C^2}} \]
In our case, the point is the origin \( (0, 0, 0) \) and the distance \( d \) is given as 5 units.

### Step 3: Set up the equation for the distance
Substituting the values into the distance formula:
\[ 5 = \frac{|2(0) + (-1)(0) + 2(0) - D|}{\sqrt{2^2 + (-1)^2 + 2^2}} \]
This simplifies to:
\[ 5 = \frac{| -D |}{\sqrt{4 + 1 + 4}} = \frac{| -D |}{\sqrt{9}} = \frac{| -D |}{3} \]

### Step 4: Solve for \( D \)
Multiplying both sides by 3 gives:
\[ 15 = | -D | \]
This implies:
\[ D = 15 \quad \text{or} \quad D = -15 \]

### Step 5: Write the equation of the plane
Using \( D = 15 \) (the positive value, as the distance is always positive), the equation of the plane can be written as:
\[ 2x - y + 2z = 15 \]

### Step 6: Vector equation of the plane
The vector equation of the plane can be expressed as:
\[ \vec{r} \cdot \vec{n} = D \]
Where \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) and \( \vec{n} = 2 \hat{i} - \hat{j} + 2 \hat{k} \). Thus, the vector equation of the plane is:
\[ \vec{r} \cdot (2 \hat{i} - \hat{j} + 2 \hat{k}) = 15 \]

### Final Answer
The vector equation of the plane is:
\[ 2x - y + 2z = 15 \]

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