`vecP` is resultant of `vecA` and `vecB. vecQ` is resultant of `vecA and -vecB` , the value of `absP^2 + absQ^2` would be

Options

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

Answer

To solve the problem, we need to find the value of \( |\vec{P}|^2 + |\vec{Q}|^2 \), where \( \vec{P} \) is the resultant of vectors \( \vec{A} \) and \( \vec{B} \), and \( \vec{Q} \) is the resultant of \( \vec{A} \) and \( -\vec{B} \).

### Step-by-Step Solution:

1. **Understanding the Resultants**:
- The resultant \( \vec{P} \) of vectors \( \vec{A} \) and \( \vec{B} \) can be expressed using the formula:
\[ |\vec{P}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta \]
where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).

2. **Finding \( |\vec{Q}|^2 \)**:
- The resultant \( \vec{Q} \) of vectors \( \vec{A} \) and \( -\vec{B} \) can be expressed as:
\[ |\vec{Q}|^2 = |\vec{A}|^2 + |-\vec{B}|^2 + 2 |\vec{A}| |-\vec{B}| \cos \phi \]
Here, \( \phi \) is the angle between \( \vec{A} \) and \( -\vec{B} \). Since \( -\vec{B} \) is in the opposite direction to \( \vec{B} \), we have \( \phi = \theta + 180^\circ \) or \( \cos \phi = -\cos \theta \). Thus, we can write:
\[ |\vec{Q}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta \]

3. **Adding \( |\vec{P}|^2 \) and \( |\vec{Q}|^2 \)**:
- Now, we add the two expressions for \( |\vec{P}|^2 \) and \( |\vec{Q}|^2 \):
\[ |\vec{P}|^2 + |\vec{Q}|^2 = (|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta) + (|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta) \]

4. **Simplifying the Expression**:
- Combining the terms:
\[ |\vec{P}|^2 + |\vec{Q}|^2 = 2 |\vec{A}|^2 + 2 |\vec{B}|^2 \]
- This can be factored as:
\[ |\vec{P}|^2 + |\vec{Q}|^2 = 2 (|\vec{A}|^2 + |\vec{B}|^2) \]

### Final Result:
Thus, the value of \( |\vec{P}|^2 + |\vec{Q}|^2 \) is:
\[ |\vec{P}|^2 + |\vec{Q}|^2 = 2 (|\vec{A}|^2 + |\vec{B}|^2) \]