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

Answer

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

### Step 1: Identify the normal vector
The normal vector \(\mathbf{n}\) is given as:
\[ \mathbf{n} = \hat{i} + 2\hat{j} - 2\hat{k} \]

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

### Step 3: Find the unit normal vector
The unit normal vector \(\mathbf{n_u}\) is given by:
\[ \mathbf{n_u} = \frac{\mathbf{n}}{|\mathbf{n}|} = \frac{\hat{i} + 2\hat{j} - 2\hat{k}}{3} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k} \]

### Step 4: Use the formula for the vector equation of the plane
The vector equation of a plane can be expressed as:
\[ \mathbf{r} \cdot \mathbf{n_u} = d \]
where \(d\) is the distance from the origin to the plane. In this case, \(d = 5\).

### Step 5: Substitute the values into the equation
Substituting the values, we have:
\[ \mathbf{r} \cdot \left(\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}\right) = 5 \]

### Step 6: Simplify the equation
Multiplying both sides by 3 to eliminate the fraction:
\[ \mathbf{r} \cdot \left(\hat{i} + 2\hat{j} - 2\hat{k}\right) = 15 \]

### Step 7: Write the final vector equation of the plane
Thus, the vector equation of the plane is:
\[ \mathbf{r} \cdot \left(\hat{i} + 2\hat{j} - 2\hat{k}\right) = 15 \]

### Summary
The required vector equation of the plane is:
\[ \mathbf{r} \cdot \left(\hat{i} + 2\hat{j} - 2\hat{k}\right) = 15 \]