The resultant of two vectors `vec(P)` and `vec(Q)` is `vec(R)`. If `vec(Q)` is doubled then the new resultant vector is perpendicular to `vec(P)`. Then magnitude of `vec(R)` is :-

Options

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

Explanation

To solve the problem step by step, we will denote the vectors and their magnitudes clearly.

### Step 1: Understand the given vectors
Let:
- \( \vec{P} \) be the vector \( \vec{P} \) with magnitude \( P \)
- \( \vec{Q} \) be the vector \( \vec{Q} \) with magnitude \( Q \)
- The resultant vector \( \vec{R} \) is given by:
\[ \vec{R} = \vec{P} + \vec{Q} \]

### Step 2: Express the new resultant when \( \vec{Q} \) is doubled
When \( \vec{Q} \) is doubled, the new resultant vector \( \vec{R}' \) becomes:
\[ \vec{R}' = \vec{P} + 2\vec{Q} \]

### Step 3: Use the condition that \( \vec{R}' \) is perpendicular to \( \vec{P} \)
Since \( \vec{R}' \) is perpendicular to \( \vec{P} \), we can use the dot product:
\[ \vec{R}' \cdot \vec{P} = 0 \]
Substituting for \( \vec{R}' \):
\[ (\vec{P} + 2\vec{Q}) \cdot \vec{P} = 0 \]

### Step 4: Expand the dot product
Expanding the dot product gives:
\[ \vec{P} \cdot \vec{P} + 2\vec{Q} \cdot \vec{P} = 0 \]
This simplifies to:
\[ P^2 + 2(\vec{Q} \cdot \vec{P}) = 0 \]
Rearranging gives:
\[ 2(\vec{Q} \cdot \vec{P}) = -P^2 \]
Thus,
\[ \vec{Q} \cdot \vec{P} = -\frac{P^2}{2} \]

### Step 5: Calculate the magnitude of the resultant vector \( \vec{R} \)
The magnitude of the resultant vector \( \vec{R} \) is given by:
\[ R^2 = P^2 + Q^2 + 2(\vec{P} \cdot \vec{Q}) \]
Substituting \( \vec{Q} \cdot \vec{P} \):
\[ R^2 = P^2 + Q^2 + 2\left(-\frac{P^2}{2}\right) \]
This simplifies to:
\[ R^2 = P^2 + Q^2 - P^2 = Q^2 \]
Taking the square root gives:
\[ R = Q \]

### Conclusion
The magnitude of the resultant vector \( \vec{R} \) is equal to the magnitude of vector \( \vec{Q} \):
\[ \boxed{Q} \]