Resultant of two vectors of magnitudes P and Q is of magnitude 'Q'. If the magnitude of `vecQ` is doubled now the angle made by new resultant with `vecP` is 

Options

A. `30^@`
B. `90^@`
C. `60^@`
D. `120^@`

Answer

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

### Step 1: Understand the Given Information
We have two vectors, \( \vec{P} \) and \( \vec{Q} \). The resultant of these two vectors is equal to the magnitude of \( \vec{Q} \). This can be expressed mathematically as:
\[ R = Q \]

### Step 2: Use the Resultant Formula
The formula for the resultant \( R \) of two vectors \( \vec{P} \) and \( \vec{Q} \) is given by:
\[ R^2 = P^2 + Q^2 + 2PQ \cos \alpha \]
where \( \alpha \) is the angle between \( \vec{P} \) and \( \vec{Q} \).

Since we know that \( R = Q \), we can substitute this into the equation:
\[ Q^2 = P^2 + Q^2 + 2PQ \cos \alpha \]

### Step 3: Simplify the Equation
By rearranging the equation, we can eliminate \( Q^2 \) from both sides:
\[ 0 = P^2 + 2PQ \cos \alpha \]
This simplifies to:
\[ P^2 = -2PQ \cos \alpha \]

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

### Step 5: Doubling the Magnitude of \( \vec{Q} \)
Now, we double the magnitude of \( \vec{Q} \). Let the new vector be \( \vec{Q'} = 2Q \). We need to find the new resultant \( R' \) and the angle \( \theta \) made by \( R' \) with \( \vec{P} \).

### Step 6: Calculate the New Resultant
Using the resultant formula again with the new \( \vec{Q'} \):
\[ R'^2 = P^2 + (2Q)^2 + 2P(2Q) \cos \alpha \]
Substituting \( \cos \alpha = -\frac{P}{2Q} \):
\[ R'^2 = P^2 + 4Q^2 - 2PQ \left(-\frac{P}{2Q}\right) \]
This simplifies to:
\[ R'^2 = P^2 + 4Q^2 + P^2 = 2P^2 + 4Q^2 \]

### Step 7: Find the Angle \( \theta \)
To find the angle \( \theta \) made by the resultant \( R' \) with \( \vec{P} \), we use the tangent of the angle:
\[ \tan \theta = \frac{2Q \sin \alpha}{R' + 2Q \cos \alpha} \]
Since \( \cos \alpha = -\frac{P}{2Q} \), we can substitute this into the equation.

### Step 8: Analyze the Result
As we calculate \( \tan \theta \), we find that the denominator approaches zero, leading to:
\[ \tan \theta \to \infty \]
This implies that:
\[ \theta = 90^\circ \]

### Final Answer
The angle made by the new resultant with \( \vec{P} \) is:
\[ \theta = 90^\circ \]