The ratio of maximum and minimum magnitude of the resultant of two vectors `vecA` and `vecB` is 3:2. The relation between A and B is

Options

A. A=5B
B. 5A=B
C. A=3B
D. A=4B

Answer

To solve the problem, we need to find the relation between the magnitudes of two vectors, \( \vec{A} \) and \( \vec{B} \), given that the ratio of the maximum and minimum magnitudes of their resultant is \( 3:2 \).

### Step-by-Step Solution:

1. **Understanding the Resultant Magnitudes**:
- The maximum magnitude of the resultant of two vectors \( \vec{A} \) and \( \vec{B} \) occurs when the vectors are in the same direction. This is given by:
\[ R_{\text{max}} = A + B \]
- The minimum magnitude of the resultant occurs when the vectors are in opposite directions. This is given by:
\[ R_{\text{min}} = |A - B| \]

2. **Setting Up the Ratio**:
- According to the problem, the ratio of the maximum to the minimum magnitude of the resultant is given as:
\[ \frac{R_{\text{max}}}{R_{\text{min}}} = \frac{3}{2} \]
- Substituting the expressions for \( R_{\text{max}} \) and \( R_{\text{min}} \):
\[ \frac{A + B}{|A - B|} = \frac{3}{2} \]

3. **Cross Multiplying**:
- Cross-multiplying to eliminate the fraction gives:
\[ 2(A + B) = 3|A - B| \]

4. **Considering Cases for \( |A - B| \)**:
- We need to consider two cases for \( |A - B| \):
- Case 1: \( A \geq B \) → \( |A - B| = A - B \)
- Case 2: \( A < B \) → \( |A - B| = B - A \)

5. **Solving Case 1**:
- For \( A \geq B \):
\[ 2(A + B) = 3(A - B) \]
Expanding both sides:
\[ 2A + 2B = 3A - 3B \]
Rearranging gives:
\[ 2A + 2B + 3B = 3A \]
\[ 2A + 5B = 3A \]
\[ 5B = 3A - 2A \]
\[ A = 5B \]

6. **Solving Case 2**:
- For \( A < B \):
\[ 2(A + B) = 3(B - A) \]
Expanding both sides:
\[ 2A + 2B = 3B - 3A \]
Rearranging gives:
\[ 2A + 3A = 3B - 2B \]
\[ 5A = B \]
This gives \( B = 5A \), which is not consistent with our earlier result.

7. **Conclusion**:
- The only valid relation from our analysis is:
\[ A = 5B \]

### Final Relation:
Thus, the relation between \( A \) and \( B \) is:
\[ A = 5B \]