If `vecx and vecy` are two vector such that `abs(vecx)=abs(vecy) and abs(vecx-vecy)=nabs(vecx+vecy)` then angle between `vecx and vecy`

Options

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

Answer

To solve the problem, we need to analyze the given conditions involving the vectors \(\vec{x}\) and \(\vec{y}\).

### Step-by-Step Solution:

1. **Given Conditions**:
We know that:
\[ |\vec{x}| = |\vec{y}| \]
Let \( |\vec{x}| = |\vec{y}| = k \).

2. **Using the Magnitude Condition**:
The second condition given is:
\[ |\vec{x} - \vec{y}| = n |\vec{x} + \vec{y}| \]
Squaring both sides, we get:
\[ |\vec{x} - \vec{y}|^2 = n^2 |\vec{x} + \vec{y}|^2 \]

3. **Expanding Both Sides**:
We can expand both sides using the properties of dot products:
\[ |\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2 \vec{x} \cdot \vec{y} \]
\[ |\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2 \vec{x} \cdot \vec{y} \]
Substituting \( |\vec{x}| = |\vec{y}| = k \):
\[ |\vec{x} - \vec{y}|^2 = k^2 + k^2 - 2 \vec{x} \cdot \vec{y} = 2k^2 - 2 \vec{x} \cdot \vec{y} \]
\[ |\vec{x} + \vec{y}|^2 = k^2 + k^2 + 2 \vec{x} \cdot \vec{y} = 2k^2 + 2 \vec{x} \cdot \vec{y} \]

4. **Setting Up the Equation**:
Now substituting these into the squared equation:
\[ 2k^2 - 2 \vec{x} \cdot \vec{y} = n^2 (2k^2 + 2 \vec{x} \cdot \vec{y}) \]
Rearranging gives:
\[ 2k^2 - 2 \vec{x} \cdot \vec{y} = 2n^2 k^2 + 2n^2 \vec{x} \cdot \vec{y} \]

5. **Combining Like Terms**:
Bringing all terms involving \(\vec{x} \cdot \vec{y}\) to one side:
\[ 2k^2 - 2n^2 k^2 = 2 \vec{x} \cdot \vec{y} (1 + n^2) \]
Simplifying gives:
\[ 2k^2(1 - n^2) = 2 \vec{x} \cdot \vec{y} (1 + n^2) \]
Dividing both sides by 2:
\[ k^2(1 - n^2) = \vec{x} \cdot \vec{y} (1 + n^2) \]

6. **Expressing \(\vec{x} \cdot \vec{y}\)**:
Recall that:
\[ \vec{x} \cdot \vec{y} = |\vec{x}| |\vec{y}| \cos \theta = k^2 \cos \theta \]
Substituting this into the equation gives:
\[ k^2(1 - n^2) = k^2 \cos \theta (1 + n^2) \]

7. **Canceling \(k^2\)**:
Assuming \(k^2 \neq 0\), we can divide both sides by \(k^2\):
\[ 1 - n^2 = \cos \theta (1 + n^2) \]

8. **Solving for \(\cos \theta\)**:
Rearranging gives:
\[ \cos \theta = \frac{1 - n^2}{1 + n^2} \]

9. **Finding the Angle \(\theta\)**:
Therefore, the angle \(\theta\) between the vectors \(\vec{x}\) and \(\vec{y}\) can be expressed as:
\[ \theta = \cos^{-1}\left(\frac{1 - n^2}{1 + n^2}\right) \]

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