Two vectors `vec(a)` and `vec(b)` are such that `|vec(a)+vec(b)|=|vec(a)-vec(b)|`. What is the angle between `vec(a)` and `vec(b)`?

Options

A. `0^(@)`
B. `90^(@)`
C. `60^(@)`
D. `180^(@)`

Explanation

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 equation**:
\[ |\vec{a} + \vec{b}| = |\vec{a} - \vec{b}| \]

2. **Square both sides**:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2 \]

3. **Expand both sides using the property of magnitudes**:
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) \]

4. **Apply the dot product**:
\[ \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \]

5. **Simplify the equation**:
The terms \(\vec{a} \cdot \vec{a}\) and \(\vec{b} \cdot \vec{b}\) cancel out on 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. **Solve for the dot product**:
\[ \vec{a} \cdot \vec{b} = 0 \]

8. **Interpret the result**:
The dot product of two vectors is given by:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Since \(\vec{a} \cdot \vec{b} = 0\), it implies:
\[ |\vec{a}| |\vec{b}| \cos \theta = 0 \]

9. **Determine the angle**:
For the product to be zero, \(\cos \theta = 0\). Therefore, the angle \(\theta\) must be:
\[ \theta = 90^\circ \]

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