If `vecP + vecQ = vecR` and `vecP - vecQ = vecS` , then `R^(2) + S^(2)` is equal to

Options

A. `P^(2) + Q^(2)`
B. `2(P^(2)-Q^(2))`
C. `2(P^(2)+Q^(2))`
D. 4PQ

Answer

To solve the problem, we start with the given equations:

1. \( \vec{P} + \vec{Q} = \vec{R} \)
2. \( \vec{P} - \vec{Q} = \vec{S} \)

We need to find the value of \( R^2 + S^2 \).

### Step 1: Square both equations

First, we square both sides of the equations:

- For the first equation:
\[ (\vec{P} + \vec{Q})^2 = \vec{R}^2 \]
Expanding this using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \):
\[ \vec{P}^2 + 2\vec{P} \cdot \vec{Q} + \vec{Q}^2 = \vec{R}^2 \]

- For the second equation:
\[ (\vec{P} - \vec{Q})^2 = \vec{S}^2 \]
Expanding this using the formula \( (a - b)^2 = a^2 - 2ab + b^2 \):
\[ \vec{P}^2 - 2\vec{P} \cdot \vec{Q} + \vec{Q}^2 = \vec{S}^2 \]

### Step 2: Add the two equations

Now, we add the two results together:
\[ (\vec{P}^2 + 2\vec{P} \cdot \vec{Q} + \vec{Q}^2) + (\vec{P}^2 - 2\vec{P} \cdot \vec{Q} + \vec{Q}^2) = \vec{R}^2 + \vec{S}^2 \]

### Step 3: Simplify the left-hand side

On the left-hand side, we can combine like terms:
\[ \vec{P}^2 + \vec{P}^2 + \vec{Q}^2 + \vec{Q}^2 + 2\vec{P} \cdot \vec{Q} - 2\vec{P} \cdot \vec{Q} = 2\vec{P}^2 + 2\vec{Q}^2 \]

So we have:
\[ 2\vec{P}^2 + 2\vec{Q}^2 = \vec{R}^2 + \vec{S}^2 \]

### Step 4: Factor out the common term

Factoring out the 2 from the left-hand side gives us:
\[ 2(\vec{P}^2 + \vec{Q}^2) = \vec{R}^2 + \vec{S}^2 \]

### Conclusion

Thus, we can conclude that:
\[ \vec{R}^2 + \vec{S}^2 = 2(\vec{P}^2 + \vec{Q}^2) \]

### Final Answer

The value of \( R^2 + S^2 \) is equal to \( 2\vec{P}^2 + 2\vec{Q}^2 \).