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
B. `(P+Q)`
C. Q
D. `(P-Q)`

Explanation

To solve the problem step by step, we will analyze the given information and apply vector addition principles.

### Step 1: Understand the Given Information
We have two vectors, \( \vec{P} \) and \( \vec{Q} \), whose resultant is \( \vec{R} \). When \( \vec{Q} \) is doubled, the new resultant vector becomes perpendicular to \( \vec{P} \).

### Step 2: Write the Expression for the Resultant
The resultant vector \( \vec{R} \) can be expressed as:
\[ \vec{R} = \vec{P} + \vec{Q} \]
We denote the magnitudes of these vectors as \( P \) and \( Q \), respectively. Thus:
\[ R = \sqrt{P^2 + Q^2 + 2PQ \cos \theta} \]
where \( \theta \) is the angle between \( \vec{P} \) and \( \vec{Q} \).

### Step 3: Analyze the Condition When \( \vec{Q} \) is Doubled
When \( \vec{Q} \) is doubled, the new vector becomes \( 2\vec{Q} \). The new resultant \( \vec{R}' \) is given by:
\[ \vec{R}' = \vec{P} + 2\vec{Q} \]
The magnitude of the new resultant is:
\[ R' = \sqrt{P^2 + (2Q)^2 + 2P(2Q) \cos \theta} \]
This simplifies to:
\[ R' = \sqrt{P^2 + 4Q^2 + 4PQ \cos \theta} \]

### Step 4: Use the Perpendicular Condition
Since \( \vec{R}' \) is perpendicular to \( \vec{P} \), the dot product \( \vec{R}' \cdot \vec{P} = 0 \). This implies:
\[ P^2 + 4Q^2 + 4PQ \cos \theta = 0 \]

### Step 5: Solve for \( \cos \theta \)
From the equation \( P^2 + 4Q^2 + 4PQ \cos \theta = 0 \), we can isolate \( \cos \theta \):
\[ 4PQ \cos \theta = - (P^2 + 4Q^2) \]
\[ \cos \theta = -\frac{P^2 + 4Q^2}{4PQ} \]

### Step 6: Substitute \( \cos \theta \) into the Resultant Equation
Now we substitute \( \cos \theta \) back into the expression for \( R \):
\[ R = \sqrt{P^2 + Q^2 + 2PQ \left(-\frac{P^2 + 4Q^2}{4PQ}\right)} \]
This simplifies to:
\[ R = \sqrt{P^2 + Q^2 - \frac{P^2 + 4Q^2}{2}} \]
\[ R = \sqrt{P^2 + Q^2 - \frac{P^2}{2} - 2Q^2} \]
\[ R = \sqrt{\frac{P^2}{2} - Q^2} \]

### Step 7: Find the Magnitude of \( \vec{R} \)
To find the magnitude of \( \vec{R} \), we need to ensure that the expression under the square root is non-negative. Given that the resultant \( R \) must be a positive quantity, we can analyze the conditions under which this holds true.

### Conclusion
From the analysis, we find that the magnitude of the resultant vector \( \vec{R} \) is:
\[ R = Q \]