If A and B are two vectors such that |A+B|=2|A-B|. The angle between vectors A and B is

Options

A. `45^(@)`
B. `60^(@)`
C. `30^(@)`
D. data insuffcient

Explanation

To solve the problem, we need to find the angle between two vectors A and B given the condition |A + B| = 2|A - B|.

### Step-by-Step Solution:

1. **Start with the Given Condition:**
We have the equation:
\[ |A + B| = 2 |A - B| \]

2. **Use the Magnitude Formulas:**
We can express the magnitudes using the formula for the magnitude of the sum and difference of two vectors:
\[ |A + B|^2 = |A|^2 + |B|^2 + 2|A||B|\cos\theta \]
\[ |A - B|^2 = |A|^2 + |B|^2 - 2|A||B|\cos\theta \]

3. **Square Both Sides of the Given Condition:**
Squaring both sides of the original equation gives:
\[ |A + B|^2 = 4 |A - B|^2 \]

4. **Substitute the Magnitude Formulas:**
Substitute the expressions for |A + B|^2 and |A - B|^2:
\[ |A|^2 + |B|^2 + 2|A||B|\cos\theta = 4(|A|^2 + |B|^2 - 2|A||B|\cos\theta) \]

5. **Expand the Right Side:**
Expanding the right side gives:
\[ |A|^2 + |B|^2 + 2|A||B|\cos\theta = 4|A|^2 + 4|B|^2 - 8|A||B|\cos\theta \]

6. **Rearranging the Equation:**
Rearranging the equation leads to:
\[ |A|^2 + |B|^2 + 2|A||B|\cos\theta - 4|A|^2 - 4|B|^2 + 8|A||B|\cos\theta = 0 \]
This simplifies to:
\[ -3|A|^2 - 3|B|^2 + 10|A||B|\cos\theta = 0 \]

7. **Factor Out Common Terms:**
Dividing through by 3 gives:
\[ -|A|^2 - |B|^2 + \frac{10}{3}|A||B|\cos\theta = 0 \]

8. **Rearranging for Cosine:**
Rearranging for \(\cos\theta\) gives:
\[ \frac{10}{3}|A||B|\cos\theta = |A|^2 + |B|^2 \]
Thus,
\[ \cos\theta = \frac{3(|A|^2 + |B|^2)}{10|A||B|} \]

9. **Substituting Values (If Needed):**
If we assume |A| = |B|, we can denote |A| = |B| = k. Then:
\[ \cos\theta = \frac{3(2k^2)}{10k^2} = \frac{6}{10} = \frac{3}{5} \]

10. **Finding the Angle:**
Finally, we find the angle \(\theta\):
\[ \theta = \cos^{-1}\left(\frac{3}{5}\right) \]

### Conclusion:
The angle between vectors A and B is:
\[ \theta \approx 53.13^\circ \]