If `veca" and "vecb` are any two vectors such `|veca+vecb|=|veca-vecb|` find the angle between `veca" and "vecb`

Answer

To solve the problem, we need to find the angle between 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**:
\[ |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \]
\[ |\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2 \vec{a} \cdot \vec{b} \]

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

5. **Cancel out \( |\vec{a}|^2 + |\vec{b}|^2 \) 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 \( 4 \) is non-zero, we can conclude**:
\[ \vec{a} \cdot \vec{b} = 0 \]

8. **Interpret the result**:
The dot product of two vectors is zero when the angle \( \theta \) between them is \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).

9. **Conclusion**:
Therefore, the angle between \( \vec{a} \) and \( \vec{b} \) is:
\[ \theta = \frac{\pi}{2} \text{ radians or } 90^\circ \]