A vector of magnitude a is turned through angle `theta` . The magnitude of change in the vector is given by:

Options

A. `|2a sin theta|`
B. `|2a sin theta //2|`
C. `|(a)/(2) sin theta|`
D. `|(a)/(2) sin theta //2|`

Answer

To find the magnitude of change in a vector of magnitude \( A \) when it is turned through an angle \( \theta \), we can follow these steps:

### Step 1: Define the initial vector
Let the initial vector \( \mathbf{A} \) be represented as:
\[ \mathbf{A} = A \hat{i} \]
This means it is directed along the x-axis with magnitude \( A \).

### Step 2: Define the new vector after rotation
When the vector is turned through an angle \( \theta \), the new vector \( \mathbf{A'} \) can be expressed in terms of its components:
\[ \mathbf{A'} = A \cos(\theta) \hat{i} + A \sin(\theta) \hat{j} \]
Here, \( A \cos(\theta) \) is the x-component and \( A \sin(\theta) \) is the y-component of the new vector.

### Step 3: Calculate the change in the vector
The change in the vector \( \Delta \mathbf{A} \) is given by:
\[ \Delta \mathbf{A} = \mathbf{A'} - \mathbf{A} \]
Substituting the expressions for \( \mathbf{A'} \) and \( \mathbf{A} \):
\[ \Delta \mathbf{A} = (A \cos(\theta) \hat{i} + A \sin(\theta) \hat{j}) - (A \hat{i}) \]
This simplifies to:
\[ \Delta \mathbf{A} = (A \cos(\theta) - A) \hat{i} + A \sin(\theta) \hat{j} \]
\[ \Delta \mathbf{A} = A (\cos(\theta) - 1) \hat{i} + A \sin(\theta) \hat{j} \]

### Step 4: Find the magnitude of the change in the vector
To find the magnitude of \( \Delta \mathbf{A} \), we use the formula for the magnitude of a vector:
\[ |\Delta \mathbf{A}| = \sqrt{(A (\cos(\theta) - 1))^2 + (A \sin(\theta))^2} \]
Expanding this:
\[ |\Delta \mathbf{A}| = \sqrt{A^2 (\cos(\theta) - 1)^2 + A^2 \sin^2(\theta)} \]
Factoring out \( A^2 \):
\[ |\Delta \mathbf{A}| = A \sqrt{(\cos(\theta) - 1)^2 + \sin^2(\theta)} \]

### Step 5: Simplify the expression
Now we simplify the expression inside the square root:
\[ |\Delta \mathbf{A}| = A \sqrt{\cos^2(\theta) - 2\cos(\theta) + 1 + \sin^2(\theta)} \]
Using the Pythagorean identity \( \cos^2(\theta) + \sin^2(\theta) = 1 \):
\[ |\Delta \mathbf{A}| = A \sqrt{1 - 2\cos(\theta) + 1} = A \sqrt{2 - 2\cos(\theta)} \]
Factoring out the 2:
\[ |\Delta \mathbf{A}| = A \sqrt{2(1 - \cos(\theta))} \]
Using the trigonometric identity \( 1 - \cos(\theta) = 2 \sin^2(\frac{\theta}{2}) \):
\[ |\Delta \mathbf{A}| = A \sqrt{2 \cdot 2 \sin^2\left(\frac{\theta}{2}\right)} = 2A \sin\left(\frac{\theta}{2}\right) \]

### Final Result
Thus, the magnitude of the change in the vector is:
\[ |\Delta \mathbf{A}| = 2A \sin\left(\frac{\theta}{2}\right) \]