If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vector, the angle between these Vector is

Options

A. `90^(@)`
B. `45^(@)`
C. `108^(@)`
D. `0^(@)`

Explanation

To solve the problem, we need to analyze the given condition mathematically. The condition states that the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors. Let's denote the two vectors as **A** and **B**.

### Step-by-step solution:

1. **Write the expressions for the magnitude of the sum and difference of two vectors:**
- The magnitude of the sum of two vectors **A** and **B** is given by:
\[ |\mathbf{A} + \mathbf{B}| = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]
- The magnitude of the difference of two vectors **A** and **B** is given by:
\[ |\mathbf{A} - \mathbf{B}| = \sqrt{A^2 + B^2 - 2AB \cos \theta} \]

2. **Set the magnitudes equal to each other:**
- According to the problem, we have:
\[ |\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}| \]
- Therefore, we can write:
\[ \sqrt{A^2 + B^2 + 2AB \cos \theta} = \sqrt{A^2 + B^2 - 2AB \cos \theta} \]

3. **Square both sides to eliminate the square roots:**
- Squaring both sides gives:
\[ A^2 + B^2 + 2AB \cos \theta = A^2 + B^2 - 2AB \cos \theta \]

4. **Simplify the equation:**
- Cancel out \(A^2 + B^2\) from both sides:
\[ 2AB \cos \theta = -2AB \cos \theta \]
- Rearranging gives:
\[ 2AB \cos \theta + 2AB \cos \theta = 0 \]
\[ 4AB \cos \theta = 0 \]

5. **Solve for \(\cos \theta\):**
- Since \(A\) and \(B\) are non-zero vectors, we can divide by \(4AB\) (which is non-zero):
\[ \cos \theta = 0 \]

6. **Determine the angle \(\theta\):**
- The cosine of an angle is zero when the angle is:
\[ \theta = 90^\circ \]

### Final Answer:
The angle between the two vectors is \(90^\circ\).