The resultant of `vec(A)+vec(B) is vec(R )_(1)`. On reversing the vector `vec(B)`, the resultant becomes `vec(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 \( R_1^2 + R_2^2 \) given the resultant vectors when adding two vectors \( \vec{A} \) and \( \vec{B} \), and then when \( \vec{B} \) is reversed.

### Step-by-step Solution:

1. **Define the Resultant Vectors**:
- The first resultant when adding vectors \( \vec{A} \) and \( \vec{B} \) is:
\[ \vec{R_1} = \vec{A} + \vec{B} \]
- The second resultant when \( \vec{B} \) is reversed (i.e., \( \vec{B} \) becomes \( -\vec{B} \)) is:
\[ \vec{R_2} = \vec{A} - \vec{B} \]

2. **Calculate the Magnitudes of the Resultant Vectors**:
- The magnitude of \( \vec{R_1} \) can be calculated using the law of cosines:
\[ R_1^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta \]
- The magnitude of \( \vec{R_2} \) is:
\[ R_2^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta \]

3. **Add the Two Resultant Squares**:
- Now, we can add \( R_1^2 \) and \( R_2^2 \):
\[ R_1^2 + R_2^2 = \left( |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta \right) + \left( |\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta \right) \]
- Simplifying this expression:
\[ R_1^2 + R_2^2 = 2 |\vec{A}|^2 + 2 |\vec{B}|^2 \]
- Thus, we can factor out the 2:
\[ R_1^2 + R_2^2 = 2 \left( |\vec{A}|^2 + |\vec{B}|^2 \right) \]

### Final Answer:
The value of \( R_1^2 + R_2^2 \) is:
\[ R_1^2 + R_2^2 = 2 \left( |\vec{A}|^2 + |\vec{B}|^2 \right) \]