The resultant of A and B is `R_(1)` On reversing the vector B , the resultant `R_(2)` what is the value of `R_(1)^(2)+ R_(2)^(2)`?

Options

A. `A^(2)+B^(2)`
B. `A^(2)-B^(2)`
C. `2(A^(2)+B^(2))`
D. `2(A^(2)-B^(2))`

Explanation

To solve the problem, we need to find the value of \( R_1^2 + R_2^2 \), where \( R_1 \) is the resultant of vectors \( A \) and \( B \), and \( R_2 \) is the resultant when vector \( B \) is reversed.

### Step-by-Step Solution:

1. **Understanding Resultants**:
- The resultant \( R_1 \) of vectors \( A \) and \( B \) can be calculated using the formula:
\[ R_1 = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]
- Here, \( A \) and \( B \) are the magnitudes of the vectors, and \( \theta \) is the angle between them.

2. **Calculating \( R_1^2 \)**:
- Squaring the resultant \( R_1 \):
\[ R_1^2 = A^2 + B^2 + 2AB \cos \theta \]

3. **Reversing Vector B**:
- When vector \( B \) is reversed, the angle between \( A \) and \( B \) becomes \( 180^\circ - \theta \).
- The cosine of \( 180^\circ - \theta \) is \( -\cos \theta \).

4. **Calculating \( R_2 \)**:
- The resultant \( R_2 \) when \( B \) is reversed is given by:
\[ R_2 = \sqrt{A^2 + B^2 + 2AB \cos(180^\circ - \theta)} = \sqrt{A^2 + B^2 - 2AB \cos \theta} \]

5. **Calculating \( R_2^2 \)**:
- Squaring the resultant \( R_2 \):
\[ R_2^2 = A^2 + B^2 - 2AB \cos \theta \]

6. **Adding \( R_1^2 \) and \( R_2^2 \)**:
- Now we can add \( R_1^2 \) and \( R_2^2 \):
\[ R_1^2 + R_2^2 = (A^2 + B^2 + 2AB \cos \theta) + (A^2 + B^2 - 2AB \cos \theta) \]
- Simplifying this expression:
\[ R_1^2 + R_2^2 = 2A^2 + 2B^2 \]
- Therefore, we can write:
\[ R_1^2 + R_2^2 = 2(A^2 + B^2) \]

### Final Answer:
\[ R_1^2 + R_2^2 = 2(A^2 + B^2) \]