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. **Calculate \( |\vec{P}|^2 \)**:
\[ |\vec{P}|^2 = |\vec{A} + \vec{B}|^2 \]
Using the formula for the magnitude of the resultant of two vectors:
\[ |\vec{P}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta \]
where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).

2. **Calculate \( |\vec{Q}|^2 \)**:
\[ |\vec{Q}|^2 = |\vec{A} - \vec{B}|^2 \]
Again using the formula for the magnitude of the resultant:
\[ |\vec{Q}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta \]
Here, the angle between \( \vec{A} \) and \( -\vec{B} \) is \( 180^\circ - \theta \), which gives us the negative cosine.

3. **Add \( |\vec{P}|^2 \) and \( |\vec{Q}|^2 \)**:
\[ |\vec{P}|^2 + |\vec{Q}|^2 = (|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta) + (|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta) \]
Simplifying this:
\[ |\vec{P}|^2 + |\vec{Q}|^2 = 2|\vec{A}|^2 + 2|\vec{B}|^2 \]

4. **Final Result**:
\[ |\vec{P}|^2 + |\vec{Q}|^2 = 2(|\vec{A}|^2 + |\vec{B}|^2) \]

### Conclusion:
The value of \( |\vec{P}|^2 + |\vec{Q}|^2 \) is \( 2(|\vec{A}|^2 + |\vec{B}|^2) \).