Two vectors of equal magnitude P are inclined at some angle such that the difference in magnitude of resultant and magnitude of either of the vectors is 0.732 times either of the magnitude of vectors. If the angle between them is increased by half of its initial value then find the magnitude of difference of the vectors

Options

A. 2P
B. `sqrt(2)p`
C. `3p`
D. `sqrt(3)p`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have two vectors of equal magnitude \( P \) inclined at an angle \( \theta \). The difference between the magnitude of the resultant vector \( R \) and the magnitude of either vector \( P \) is given as \( 0.732P \).

### Step 2: Write the Expression for the Resultant
The formula for the resultant \( R \) of two vectors \( A \) and \( B \) (where \( A = B = P \)) inclined at an angle \( \theta \) is given by:
\[ R = \sqrt{P^2 + P^2 + 2P \cdot P \cdot \cos(\theta)} \]
This simplifies to:
\[ R = \sqrt{2P^2(1 + \cos(\theta))} = P\sqrt{2(1 + \cos(\theta))} \]

### Step 3: Set Up the Equation
According to the problem, we have:
\[ R - P = 0.732P \]
This can be rearranged to:
\[ R = P + 0.732P = 1.732P \]

### Step 4: Equate the Two Expressions for R
Now we can equate the two expressions for \( R \):
\[ P\sqrt{2(1 + \cos(\theta))} = 1.732P \]
Dividing both sides by \( P \) (assuming \( P \neq 0 \)):
\[ \sqrt{2(1 + \cos(\theta))} = 1.732 \]

### Step 5: Square Both Sides
Squaring both sides gives:
\[ 2(1 + \cos(\theta)) = (1.732)^2 \]
Calculating \( (1.732)^2 \):
\[ (1.732)^2 \approx 3 \]
Thus, we have:
\[ 2(1 + \cos(\theta)) = 3 \]

### Step 6: Solve for cos(θ)
Dividing both sides by 2:
\[ 1 + \cos(\theta) = 1.5 \]
Subtracting 1 from both sides:
\[ \cos(\theta) = 0.5 \]
This implies:
\[ \theta = 60^\circ \]

### Step 7: Increase the Angle
The problem states that the angle is increased by half of its initial value. Therefore, the new angle \( \phi \) is:
\[ \phi = \theta + \frac{\theta}{2} = 60^\circ + 30^\circ = 90^\circ \]

### Step 8: Find the Magnitude of the Difference of the Vectors
The magnitude of the difference of two vectors \( A \) and \( B \) when they are at \( 90^\circ \) to each other is given by:
\[ |A - B| = \sqrt{A^2 + B^2} \]
Since \( A = B = P \):
\[ |A - B| = \sqrt{P^2 + P^2} = \sqrt{2P^2} = P\sqrt{2} \]

### Final Answer
Thus, the magnitude of the difference of the vectors is:
\[ P\sqrt{2} \]

---