Two vectors `vec(A)` and `vec(B)` are such that `|vec(A)+vec(B)|=|vec(A)-vec(B)|` then what is the angle between `vec(A)`and `vec(B)` :-

Options

A. `90^(@)`
B. `60^(@)`
C. `75^(@)`
D. `45^(@)`

Answer

To solve the problem, we need to find the angle between the two vectors \(\vec{A}\) and \(\vec{B}\) given that \(|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|\).

### Step-by-Step Solution:

1. **Start with the given condition**:
\[ |\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \]

2. **Square both sides** to eliminate the square root:
\[ |\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2 \]

3. **Use the formula for the magnitude of vectors**:
\[ |\vec{A} + \vec{B}|^2 = (\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = \vec{A} \cdot \vec{A} + 2 \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{B} = A^2 + 2 \vec{A} \cdot \vec{B} + B^2 \]
\[ |\vec{A} - \vec{B}|^2 = (\vec{A} - \vec{B}) \cdot (\vec{A} - \vec{B}) = \vec{A} \cdot \vec{A} - 2 \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{B} = A^2 - 2 \vec{A} \cdot \vec{B} + B^2 \]

4. **Set the two expressions equal to each other**:
\[ A^2 + 2 \vec{A} \cdot \vec{B} + B^2 = A^2 - 2 \vec{A} \cdot \vec{B} + B^2 \]

5. **Cancel \(A^2\) and \(B^2\) from both sides**:
\[ 2 \vec{A} \cdot \vec{B} = -2 \vec{A} \cdot \vec{B} \]

6. **Combine like terms**:
\[ 2 \vec{A} \cdot \vec{B} + 2 \vec{A} \cdot \vec{B} = 0 \]
\[ 4 \vec{A} \cdot \vec{B} = 0 \]

7. **Since \(A\) and \(B\) cannot be zero**, we conclude:
\[ \vec{A} \cdot \vec{B} = 0 \]

8. **Use the dot product to find the angle**:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta = 0 \]
This implies:
\[ \cos \theta = 0 \]

9. **Determine the angle**:
\[ \theta = 90^\circ \]

### Final Answer:
The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).

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