If `vec(A),vec(B) & vec(A)+vec(B)` are three non`-` zero vector. Such that `vec(A)+vec(B)` is perpendicular to `vec(B)` then which of one is correct `:`

Options

A. `AgeB`
B. `Age(B)/(sqrt(2))`
C. `AgtB`
D. `Agt(B)/(sqrt(2))`

Explanation

To solve the problem, we need to analyze the given information about the vectors \( \vec{A} \), \( \vec{B} \), and \( \vec{A} + \vec{B} \). We know that \( \vec{A} + \vec{B} \) is perpendicular to \( \vec{B} \).

### Step-by-Step Solution:

1. **Understanding Perpendicular Vectors**:
Since \( \vec{A} + \vec{B} \) is perpendicular to \( \vec{B} \), we can use the property of dot products. For two vectors \( \vec{X} \) and \( \vec{Y} \) to be perpendicular, their dot product must equal zero:
\[ (\vec{A} + \vec{B}) \cdot \vec{B} = 0 \]

2. **Expanding the Dot Product**:
We can expand the dot product:
\[ \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{B} = 0 \]
Here, \( \vec{B} \cdot \vec{B} \) is the magnitude of \( \vec{B} \) squared, denoted as \( |\vec{B}|^2 \).

3. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ \vec{A} \cdot \vec{B} = -|\vec{B}|^2 \]

4. **Using the Dot Product Formula**:
The dot product \( \vec{A} \cdot \vec{B} \) can also be expressed in terms of the magnitudes and the cosine of the angle \( \theta \) between them:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
Thus, we can substitute this into our equation:
\[ |\vec{A}| |\vec{B}| \cos \theta = -|\vec{B}|^2 \]

5. **Dividing by \( |\vec{B}| \)**:
Assuming \( |\vec{B}| \neq 0 \) (since \( \vec{B} \) is a non-zero vector), we can divide both sides by \( |\vec{B}| \):
\[ |\vec{A}| \cos \theta = -|\vec{B}| \]

6. **Analyzing the Equation**:
Since \( |\vec{A}| \) and \( |\vec{B}| \) are both positive (as they are magnitudes), \( \cos \theta \) must be negative. This implies that the angle \( \theta \) between \( \vec{A} \) and \( \vec{B} \) is greater than 90 degrees but less than 180 degrees.

7. **Conclusion**:
The equation \( |\vec{A}| \cos \theta = -|\vec{B}| \) indicates that \( |\vec{A}| \) must be greater than or equal to \( |\vec{B}| \) in order for the equation to hold true, as \( \cos \theta \) is negative. Therefore, we conclude:
\[ |\vec{A}| \geq |\vec{B}| \]

### Final Answer:
Thus, the correct option is that \( |\vec{A}| \geq |\vec{B}| \).