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 analyze the two scenarios involving the vectors \(\vec{A}\) and \(\vec{B}\).

### Step 1: Understand the Resultant of Two Vectors
The magnitude of the resultant vector when two vectors \(\vec{A}\) and \(\vec{B}\) are added is given by the formula:
\[ P = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]
where \(\theta\) is the angle between the two vectors.

### Step 2: Resultant When \(\vec{B}\) is Reversed
When vector \(\vec{B}\) is reversed, it becomes \(-\vec{B}\). The new resultant vector \(Q\) can be expressed as:
\[ Q = \sqrt{A^2 + B^2 - 2AB \cos(\theta)} \]
This is because reversing \(\vec{B}\) changes the sign of the cosine term in the formula.

### Step 3: Calculate \(P^2\) and \(Q^2\)
Now, we can calculate \(P^2\) and \(Q^2\):
1. For \(P^2\):
\[ P^2 = A^2 + B^2 + 2AB \cos(\theta) \]
2. For \(Q^2\):
\[ Q^2 = A^2 + B^2 - 2AB \cos(\theta) \]

### Step 4: Add \(P^2\) and \(Q^2\)
Now, we add \(P^2\) and \(Q^2\):
\[ P^2 + Q^2 = (A^2 + B^2 + 2AB \cos(\theta)) + (A^2 + B^2 - 2AB \cos(\theta)) \]
The \(2AB \cos(\theta)\) terms cancel each other:
\[ P^2 + Q^2 = 2A^2 + 2B^2 \]

### Step 5: Factor the Result
We can factor out the 2:
\[ P^2 + Q^2 = 2(A^2 + B^2) \]

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