Two vectors ` vec A` and `vec B` have equal magnitudes. If magnitude of ` vec A-vec B` is equal to (n) times the magnitude of ` vec A-vec B`, then angle between ` vec A` and ` vec B` is ?

Options

A. ` cos^(-1) ((n-1)/(n-1))`
B. ` cos^(-1) ((n^2-1)/(n^2+1))`
C. ` cos^(-1) ((n-1)/(n-1))`
D. ` cos^(-1) ((n^2-1)/(n^2+1))`

Explanation

To solve the problem, we need to find the angle between two vectors \( \vec{A} \) and \( \vec{B} \) given that they have equal magnitudes and the magnitude of \( \vec{A} + \vec{B} \) is equal to \( n \) times the magnitude of \( \vec{A} - \vec{B} \).

### Step-by-Step Solution:

1. **Given Information**:
- Magnitude of \( \vec{A} \) is equal to the magnitude of \( \vec{B} \).
- Let \( |\vec{A}| = |\vec{B}| = A \).
- We have the equation:
\[ |\vec{A} + \vec{B}| = n |\vec{A} - \vec{B}| \]

2. **Express Magnitudes**:
- The magnitude of \( \vec{A} + \vec{B} \) can be expressed using the formula:
\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}||\vec{B}| \cos \theta} \]
- Since \( |\vec{A}| = |\vec{B}| = A \), this becomes:
\[ |\vec{A} + \vec{B}| = \sqrt{A^2 + A^2 + 2A^2 \cos \theta} = \sqrt{2A^2(1 + \cos \theta)} = A\sqrt{2(1 + \cos \theta)} \]

3. **Magnitude of \( \vec{A} - \vec{B} \)**:
- Similarly, the magnitude of \( \vec{A} - \vec{B} \) is given by:
\[ |\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}||\vec{B}| \cos \theta} \]
- This becomes:
\[ |\vec{A} - \vec{B}| = \sqrt{A^2 + A^2 - 2A^2 \cos \theta} = \sqrt{2A^2(1 - \cos \theta)} = A\sqrt{2(1 - \cos \theta)} \]

4. **Substituting into the Equation**:
- Now substituting these magnitudes into the original equation:
\[ A\sqrt{2(1 + \cos \theta)} = n \cdot A\sqrt{2(1 - \cos \theta)} \]
- Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \sqrt{2(1 + \cos \theta)} = n \sqrt{2(1 - \cos \theta)} \]

5. **Squaring Both Sides**:
- Squaring both sides gives:
\[ 2(1 + \cos \theta) = n^2 \cdot 2(1 - \cos \theta) \]
- Simplifying this:
\[ 1 + \cos \theta = n^2(1 - \cos \theta) \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ 1 + \cos \theta = n^2 - n^2 \cos \theta \]
- Bringing all terms involving \( \cos \theta \) to one side:
\[ \cos \theta + n^2 \cos \theta = n^2 - 1 \]
- Factoring out \( \cos \theta \):
\[ \cos \theta (1 + n^2) = n^2 - 1 \]

7. **Finding \( \cos \theta \)**:
- Finally, we can express \( \cos \theta \):
\[ \cos \theta = \frac{n^2 - 1}{1 + n^2} \]

8. **Finding the Angle**:
- The angle \( \theta \) can be found using the inverse cosine function:
\[ \theta = \cos^{-1}\left(\frac{n^2 - 1}{1 + n^2}\right) \]

### Final Answer:
The angle \( \theta \) between the vectors \( \vec{A} \) and \( \vec{B} \) is given by:
\[ \theta = \cos^{-1}\left(\frac{n^2 - 1}{1 + n^2}\right) \]