What are the direction ratios of normal to the plane `2x-y+2z+1=0`?

Options

A. `(: 2,1,2 :)`
B. `(: 1,-1/2,1 :)`
C. `(: 1,-2,1 :)`
D. None of the above

Explanation

To find the direction ratios of the normal to the plane given by the equation \(2x - y + 2z + 1 = 0\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the General Form of the Plane Equation**:
The general equation of a plane can be expressed as:
\[ ax + by + cz + d = 0 \]
where \(a\), \(b\), and \(c\) are the coefficients of \(x\), \(y\), and \(z\) respectively, and \(d\) is a constant.

2. **Compare with the Given Plane Equation**:
The given equation is:
\[ 2x - y + 2z + 1 = 0 \]
Here, we can identify the coefficients:
- \(a = 2\)
- \(b = -1\)
- \(c = 2\)

3. **Determine the Direction Ratios**:
The direction ratios of the normal to the plane are given by the coefficients \(a\), \(b\), and \(c\). Therefore, the direction ratios are:
\[ (2, -1, 2) \]

4. **Simplify the Direction Ratios (if necessary)**:
Direction ratios can be represented in a simplified form. We can divide each component by 2:
\[ \left(\frac{2}{2}, \frac{-1}{2}, \frac{2}{2}\right) = (1, -\frac{1}{2}, 1) \]

5. **Final Answer**:
The direction ratios of the normal to the plane \(2x - y + 2z + 1 = 0\) are:
\[ (1, -\frac{1}{2}, 1) \]