If the angle between two vectors of equal magnitude P is `theta` the magnitude of the difference of the vectors is

Options

A. `2P cos theta/2`
B. `2P sin theta/2`
C. `P cos theta/2`
D. `P sin theta/2`

Answer

To find the magnitude of the difference between two vectors of equal magnitude \( P \) that form an angle \( \theta \) with each other, we can follow these steps:

### Step 1: Understand the Vectors
Let the two vectors be represented as \( \vec{A} \) and \( \vec{B} \). Both vectors have the same magnitude \( P \), and the angle between them is \( \theta \).

### Step 2: Use the Formula for the Magnitude of the Difference
The magnitude of the difference of two vectors \( \vec{A} - \vec{B} \) can be calculated using the formula:
\[ |\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos(\theta)} \]
Since both vectors have the same magnitude \( P \), we can substitute \( |\vec{A}| = P \) and \( |\vec{B}| = P \).

### Step 3: Substitute the Values
Substituting the values into the formula gives:
\[ |\vec{A} - \vec{B}| = \sqrt{P^2 + P^2 - 2P \cdot P \cos(\theta)} \]
This simplifies to:
\[ |\vec{A} - \vec{B}| = \sqrt{2P^2 - 2P^2 \cos(\theta)} \]

### Step 4: Factor Out Common Terms
We can factor out \( 2P^2 \) from the square root:
\[ |\vec{A} - \vec{B}| = \sqrt{2P^2(1 - \cos(\theta))} \]
This can be simplified further:
\[ |\vec{A} - \vec{B}| = P\sqrt{2(1 - \cos(\theta))} \]

### Step 5: Use the Trigonometric Identity
Using the trigonometric identity \( 1 - \cos(\theta) = 2\sin^2(\frac{\theta}{2}) \), we can substitute:
\[ |\vec{A} - \vec{B}| = P\sqrt{2 \cdot 2\sin^2\left(\frac{\theta}{2}\right)} \]
This simplifies to:
\[ |\vec{A} - \vec{B}| = P \cdot 2\sin\left(\frac{\theta}{2}\right) \]

### Final Result
Thus, the magnitude of the difference of the vectors is:
\[ |\vec{A} - \vec{B}| = 2P\sin\left(\frac{\theta}{2}\right) \]

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