A class XII student appearing for a competitive examination was asked to attempt the following questions
Let `veca,vecb` and `vecc` be three non zero vectors.
if `veca` and `vecb` are unit vectors and Q be the angle between them then `abs(veca-vecb)` is

Options

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

Answer

To solve the problem, we need to find the magnitude of the vector difference \( |\vec{a} - \vec{b}| \) given that \( \vec{a} \) and \( \vec{b} \) are unit vectors and \( \theta \) is the angle between them.

### Step-by-Step Solution:

1. **Understanding the Magnitude of the Difference of Two Vectors**:
We start with the formula for the magnitude of the difference of two vectors:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 |\vec{a}| |\vec{b}| \cos \theta \]

2. **Substituting the Values**:
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we have:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]
Substituting these values into the equation gives:
\[ |\vec{a} - \vec{b}|^2 = 1^2 + 1^2 - 2 \cdot 1 \cdot 1 \cdot \cos \theta \]
Simplifying this, we get:
\[ |\vec{a} - \vec{b}|^2 = 1 + 1 - 2 \cos \theta = 2 - 2 \cos \theta \]

3. **Factoring Out the Common Terms**:
We can factor out the 2 from the right-hand side:
\[ |\vec{a} - \vec{b}|^2 = 2(1 - \cos \theta) \]

4. **Using the Trigonometric Identity**:
We know from trigonometric identities that:
\[ 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \]
Substituting this into our equation gives:
\[ |\vec{a} - \vec{b}|^2 = 2 \cdot 2 \sin^2\left(\frac{\theta}{2}\right) = 4 \sin^2\left(\frac{\theta}{2}\right) \]

5. **Taking the Square Root**:
Finally, we take the square root of both sides to find the magnitude:
\[ |\vec{a} - \vec{b}| = \sqrt{4 \sin^2\left(\frac{\theta}{2}\right)} = 2 \sin\left(\frac{\theta}{2}\right) \]

### Final Answer:
Thus, the magnitude of the vector difference \( |\vec{a} - \vec{b}| \) is:
\[ |\vec{a} - \vec{b}| = 2 \sin\left(\frac{\theta}{2}\right) \]