Find th equation of the plane which is at distance of 8 units from origin and the perpendicular vector from origin to this plane is `(2hati+hatj-2hatk)`.

Answer

To find the equation of the plane that is at a distance of 8 units from the origin and has a normal vector given by \( \vec{n} = 2\hat{i} + \hat{j} - 2\hat{k} \), we can follow these steps:

### Step 1: Understand the formula for the equation of a plane
The general equation of a plane can be expressed as:
\[ \vec{r} \cdot \hat{n} = d \]
where \( \vec{r} \) is the position vector of any point on the plane, \( \hat{n} \) is the unit normal vector to the plane, and \( d \) is the perpendicular distance from the origin to the plane.

### Step 2: Identify the given values
From the problem, we have:
- Distance \( d = 8 \) units
- Normal vector \( \vec{n} = 2\hat{i} + \hat{j} - 2\hat{k} \)

### Step 3: Calculate the magnitude of the normal vector
To find the unit normal vector \( \hat{n} \), we first need to calculate the magnitude of \( \vec{n} \):
\[ |\vec{n}| = \sqrt{(2)^2 + (1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 4: Find the unit normal vector
Now, we can find the unit normal vector \( \hat{n} \):
\[ \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{2\hat{i} + \hat{j} - 2\hat{k}}{3} = \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k} \]

### Step 5: Substitute into the plane equation
Now, substituting \( \hat{n} \) and \( d \) into the plane equation:
\[ \vec{r} \cdot \left(\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k}\right) = 8 \]

### Step 6: Multiply both sides by 3 to eliminate the fraction
Multiplying both sides by 3 gives:
\[ \vec{r} \cdot (2\hat{i} + \hat{j} - 2\hat{k}) = 24 \]

### Step 7: Write the final equation of the plane
The equation of the plane can thus be expressed as:
\[ 2x + y - 2z = 24 \]
where \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \).