Find the equation of a plane which is a distance of 2 units from origin and the d.r's of perpendicular vectors are `2,-1,2`.

Answer

To find the equation of a plane that is a distance of 2 units from the origin and has direction ratios of the normal vector as \(2, -1, 2\), we can follow these steps:

### Step 1: Identify the normal vector
The direction ratios of the normal vector to the plane are given as \(2, -1, 2\). We can denote the normal vector \(\mathbf{n}\) as:
\[ \mathbf{n} = 2\mathbf{i} - 1\mathbf{j} + 2\mathbf{k} \]

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

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

### Step 4: Use the distance formula for the plane
The equation of a plane can be expressed as:
\[ \mathbf{r} \cdot \mathbf{n} = d \]
where \(d\) is the distance from the origin to the plane. The distance from the origin to the plane is given as \(2\) units. Therefore, we have:
\[ \mathbf{r} \cdot (2\mathbf{i} - 1\mathbf{j} + 2\mathbf{k}) = 2 \cdot 3 = 6 \]

### Step 5: Write the equation of the plane
The equation of the plane can be written as:
\[ 2x - y + 2z = 6 \]

### Final Answer
Thus, the equation of the plane is:
\[ 2x - y + 2z = 6 \]

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