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

Options

A. `P+Q`
B. P
C. `P-Q`
D. Q

Explanation

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

### Step 1: Understand the vectors and their resultant
We have two vectors, \(\vec{P}\) and \(\vec{Q}\), whose resultant is \(\vec{R}\). The relationship can be expressed as:
\[ \vec{R} = \vec{P} + \vec{Q} \]

### Step 2: Doubling the magnitude of vector \(\vec{Q}\)
If the magnitude of \(\vec{Q}\) is doubled, we denote the new vector as \(2\vec{Q}\). The new resultant vector, which we will call \(\vec{R_1}\), can be expressed as:
\[ \vec{R_1} = \vec{P} + 2\vec{Q} \]

### Step 3: Condition of perpendicularity
According to the problem, the new resultant \(\vec{R_1}\) is perpendicular to \(\vec{P}\). This means that the dot product of \(\vec{R_1}\) and \(\vec{P}\) is zero:
\[ \vec{R_1} \cdot \vec{P} = 0 \]

### Step 4: Express the dot product
Substituting \(\vec{R_1}\) into the dot product condition, we have:
\[ (\vec{P} + 2\vec{Q}) \cdot \vec{P} = 0 \]
Expanding this gives:
\[ \vec{P} \cdot \vec{P} + 2\vec{Q} \cdot \vec{P} = 0 \]
Letting \(P = |\vec{P}|\) and \(Q = |\vec{Q}|\), we can express this as:
\[ P^2 + 2Q \cos(\theta) = 0 \]
where \(\theta\) is the angle between \(\vec{P}\) and \(\vec{Q}\).

### Step 5: Solve for \(Q\)
Rearranging the equation gives:
\[ 2Q \cos(\theta) = -P^2 \]
Thus,
\[ Q \cos(\theta) = -\frac{P^2}{2} \]

### Step 6: Find the magnitude of the resultant \(\vec{R}\)
The magnitude of the resultant vector \(\vec{R}\) can be found using the formula:
\[ R^2 = P^2 + Q^2 + 2Q \cos(\theta) \]
Substituting \(2Q \cos(\theta) = -P^2\) into the equation yields:
\[ R^2 = P^2 + Q^2 - P^2 \]
This simplifies to:
\[ R^2 = Q^2 \]
Taking the square root gives:
\[ R = Q \]

### Final Answer
The magnitude of \(\vec{R}\) is equal to the magnitude of \(\vec{Q}\):
\[ \boxed{|\vec{Q}|} \]