If `vec(A)` is perpendicular to `vec(B)`, then

Options

A. `vec(A)xxvec(B)=0`
B. `vec(A).[vec(A)+vec(B)]=A^(2)`
C. `vec(A).vec(B)=AB`
D. `vec(A).[vec(A)+vec(B)]=A^(2)+AB`

Explanation

To solve the problem, we need to analyze the implications of the vectors \( \vec{A} \) and \( \vec{B} \) being perpendicular to each other.

### Step-by-Step Solution:

1. **Understanding Perpendicular Vectors**:
- When two vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular, their dot product is zero. This can be expressed mathematically as:
\[ \vec{A} \cdot \vec{B} = 0 \]

2. **Evaluating the Options**:
- We are given several options to determine which statement is true when \( \vec{A} \) is perpendicular to \( \vec{B} \).

3. **Option 1: \( \vec{A} \times \vec{B} = 0 \)**:
- The cross product of two vectors is zero only when they are parallel or when one of the vectors is a null vector. Since \( \vec{A} \) and \( \vec{B} \) are perpendicular, this option is **incorrect**.

4. **Option 2: \( \vec{A} \cdot (\vec{A} + \vec{B}) = \vec{A}^2 \)**:
- We can expand this using the distributive property of the dot product:
\[ \vec{A} \cdot (\vec{A} + \vec{B}) = \vec{A} \cdot \vec{A} + \vec{A} \cdot \vec{B} \]
- Since \( \vec{A} \cdot \vec{B} = 0 \), we have:
\[ \vec{A} \cdot \vec{A} + 0 = \vec{A}^2 \]
- Therefore, this option is **correct**.

5. **Option 3: \( \vec{A} \cdot \vec{B} = \vec{A} \vec{B} \)**:
- This statement is incorrect because \( \vec{A} \cdot \vec{B} = 0 \) when \( \vec{A} \) is perpendicular to \( \vec{B} \).

6. **Option 4: \( \vec{A} \cdot (\vec{A} + \vec{B}) \neq \vec{A}^2 + \vec{A} \cdot \vec{B} \)**:
- This statement is also incorrect because we have already shown that \( \vec{A} \cdot (\vec{A} + \vec{B}) = \vec{A}^2 \) and \( \vec{A} \cdot \vec{B} = 0 \). Thus, this option is **wrong**.

### Conclusion:
The only correct statement when \( \vec{A} \) is perpendicular to \( \vec{B} \) is **Option 2**:
\[ \vec{A} \cdot (\vec{A} + \vec{B}) = \vec{A}^2 \]