The resultant of two vectors `vecP andvecQ is vecR`. If the magnitude of `vecQ` is doudled, the new resultant becomes perpendicuar to `vecP`. Then the magnitude of `vecR` is :

Options

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

Explanation

To solve the problem, we need to analyze the situation step by step.

### Step 1: Understand the Given Information
We have two vectors, **P** and **Q**, whose resultant is **R**. The problem states that if the magnitude of vector **Q** is doubled, the new resultant becomes perpendicular to vector **P**.

### Step 2: Write the Expression for the Resultant
The magnitude of the resultant vector **R** when combining vectors **P** and **Q** can be expressed using the formula:
\[ R^2 = P^2 + Q^2 + 2PQ \cos(\theta) \]
where \(\theta\) is the angle between vectors **P** and **Q**.

### Step 3: Modify the Expression for the New Resultant
When the magnitude of vector **Q** is doubled, we denote the new vector as **Q' = 2Q**. The new resultant vector **R'** can be expressed as:
\[ R' = P + Q' = P + 2Q \]

### Step 4: Condition for Perpendicular Vectors
Since the new resultant **R'** is perpendicular to vector **P**, we can use the dot product condition:
\[ R' \cdot P = 0 \]
This implies:
\[ (P + 2Q) \cdot P = 0 \]
Expanding this gives:
\[ P \cdot P + 2Q \cdot P = 0 \]
which simplifies to:
\[ P^2 + 2Q \cos(\theta) = 0 \]

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

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

### Step 7: Find the Magnitude of the Resultant
Taking the square root of both sides gives us:
\[ R = |Q| \]

### Final Answer
Thus, the magnitude of the resultant vector **R** is equal to the magnitude of vector **Q**.

### Conclusion
The correct answer is:
\[ R = |Q| \]