If `vec a` and `vec b` are two unit vectors and `theta` is the angle between them, then the unit vector along the angular bisector of `vec a` and `vec b` will be given by

Options

A. a) `( vec a- vec b)/(cos(theta//2))`
B. b) `( vec a+ vec b)/(2cos(theta//2))`
C. c) `( vec a- vec b)/(2cos(theta//2))`
D. d) None of these

Explanation

To find the unit vector along the angular bisector of two unit vectors \(\vec{a}\) and \(\vec{b}\) with an angle \(\theta\) between them, we can follow these steps:

### Step-by-step Solution:

1. **Understand the Given Information**:
We have two unit vectors \(\vec{a}\) and \(\vec{b}\), which means \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\). The angle between them is \(\theta\).

2. **Find the Resultant Vector**:
The vector along the angular bisector can be represented as the sum of the two vectors:
\[ \vec{R} = \vec{a} + \vec{b} \]

3. **Calculate the Magnitude of the Resultant Vector**:
The magnitude of \(\vec{R}\) can be calculated using the cosine rule:
\[ |\vec{R}| = |\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\theta} \]
Since \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\):
\[ |\vec{R}| = \sqrt{1^2 + 1^2 + 2 \cdot 1 \cdot 1 \cdot \cos\theta} = \sqrt{2 + 2\cos\theta} = \sqrt{2(1 + \cos\theta)} = \sqrt{2(2\cos^2(\theta/2))} = 2\cos(\theta/2) \]

4. **Find the Unit Vector Along the Bisector**:
The unit vector along the bisector is given by:
\[ \hat{u} = \frac{\vec{R}}{|\vec{R}|} = \frac{\vec{a} + \vec{b}}{|\vec{R}|} \]
Substituting the magnitude we found:
\[ \hat{u} = \frac{\vec{a} + \vec{b}}{2\cos(\theta/2)} \]

5. **Final Expression**:
Therefore, the unit vector along the angular bisector of \(\vec{a}\) and \(\vec{b}\) is:
\[ \hat{u} = \frac{\vec{a} + \vec{b}}{2\cos(\theta/2)} \]