A class XII student appearing for a competitive examination was asked to attempt the following questions
Let `veca,vecb` and `vecc` be three non zero vectors.
If `veca` and ?`vecb` are such that `|veca+ vecb | = |veca − vecb|` then

Options

A. `vecabotvecb`
B. `vec||vecb`
C. `veca=vecb`
D. None of these

Answer

To solve the problem, we need to show that if \(|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|\), then the vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other.

### 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 formula \(|\vec{x}|^2 = \vec{x} \cdot \vec{x}\):**
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) \]

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

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

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

7. **Simplify the equation:**
\[ 2 \vec{a} \cdot \vec{b} = -2 \vec{a} \cdot \vec{b} \]

8. **Combine like terms:**
\[ 4 \vec{a} \cdot \vec{b} = 0 \]

9. **Conclude that:**
\[ \vec{a} \cdot \vec{b} = 0 \]

Since the dot product of \(\vec{a}\) and \(\vec{b}\) is zero, this means that the vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular to each other.

### Final Answer:
\(\vec{a}\) is perpendicular to \(\vec{b}\).