IF `barx and bary` non zero linearly independent vectors such that `|barx + bary|=|barx -bary|` then

Options

A. `barx and bary` are parallel
B. `barx and bary` are perpendicular
C. angle between `barx and bary is pi/4`
D. None of these

Answer

To solve the problem, we need to find the angle between the two non-zero linearly independent vectors \( \vec{x} \) and \( \vec{y} \) given that \( |\vec{x} + \vec{y}| = |\vec{x} - \vec{y}| \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ |\vec{x} + \vec{y}| = |\vec{x} - \vec{y}| \]

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

3. **Expand both sides using the formula \( |\vec{a}|^2 = \vec{a} \cdot \vec{a} \):**
\[ (\vec{x} + \vec{y}) \cdot (\vec{x} + \vec{y}) = (\vec{x} - \vec{y}) \cdot (\vec{x} - \vec{y}) \]

4. **Calculate the left-hand side:**
\[ \vec{x} \cdot \vec{x} + 2\vec{x} \cdot \vec{y} + \vec{y} \cdot \vec{y} = |\vec{x}|^2 + 2\vec{x} \cdot \vec{y} + |\vec{y}|^2 \]

5. **Calculate the right-hand side:**
\[ \vec{x} \cdot \vec{x} - 2\vec{x} \cdot \vec{y} + \vec{y} \cdot \vec{y} = |\vec{x}|^2 - 2\vec{x} \cdot \vec{y} + |\vec{y}|^2 \]

6. **Set the two expanded forms equal to each other:**
\[ |\vec{x}|^2 + 2\vec{x} \cdot \vec{y} + |\vec{y}|^2 = |\vec{x}|^2 - 2\vec{x} \cdot \vec{y} + |\vec{y}|^2 \]

7. **Simplify the equation:**
\[ 2\vec{x} \cdot \vec{y} + 2\vec{x} \cdot \vec{y} = 0 \]
\[ 4\vec{x} \cdot \vec{y} = 0 \]

8. **Since \( \vec{x} \) and \( \vec{y} \) are non-zero vectors, we can conclude:**
\[ \vec{x} \cdot \vec{y} = 0 \]

9. **Interpret the result:**
The dot product being zero implies that the angle \( \theta \) between the vectors \( \vec{x} \) and \( \vec{y} \) is:
\[ \cos \theta = 0 \implies \theta = \frac{\pi}{2} \]

### Conclusion:
The angle between the vectors \( \vec{x} \) and \( \vec{y} \) is \( \frac{\pi}{2} \) radians, which means they are perpendicular.

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