Resultant of two vectors `vecA and vecB` is of magnitude P, If `vecB` is reversed, then resultant is of magnitude Q. What is the value of `P^(2) + Q^(2)` ?

Options

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

Explanation

To solve the problem, we need to find the value of \( P^2 + Q^2 \) based on the given conditions about the vectors \( \vec{A} \) and \( \vec{B} \).

### Step-by-Step Solution:

1. **Define the Vectors**: Let the magnitudes of the vectors be \( A \) and \( B \), and let \( \theta \) be the angle between them.

2. **Resultant Magnitude when \( \vec{B} \) is not reversed**: The magnitude of the resultant vector \( \vec{R} \) when \( \vec{B} \) is not reversed is given by:
\[ P = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]
Squaring both sides, we have:
\[ P^2 = A^2 + B^2 + 2AB \cos \theta \quad \text{(Equation 1)} \]

3. **Resultant Magnitude when \( \vec{B} \) is reversed**: When \( \vec{B} \) is reversed, the angle between \( \vec{A} \) and \( \vec{B} \) becomes \( 180^\circ + \theta \). The magnitude of the resultant vector \( Q \) is given by:
\[ Q = \sqrt{A^2 + B^2 + 2AB \cos(180^\circ + \theta)} \]
Since \( \cos(180^\circ + \theta) = -\cos \theta \), we can rewrite this as:
\[ Q = \sqrt{A^2 + B^2 - 2AB \cos \theta} \]
Squaring both sides, we have:
\[ Q^2 = A^2 + B^2 - 2AB \cos \theta \quad \text{(Equation 2)} \]

4. **Add the Two Equations**: Now, we add Equation 1 and Equation 2:
\[ P^2 + Q^2 = (A^2 + B^2 + 2AB \cos \theta) + (A^2 + B^2 - 2AB \cos \theta) \]
The \( 2AB \cos \theta \) terms cancel out:
\[ P^2 + Q^2 = 2A^2 + 2B^2 \]
Simplifying this gives:
\[ P^2 + Q^2 = 2(A^2 + B^2) \]

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