If `vec(e_(1))` and `vec(e_(2))` are two unit vectors and `theta` is the angle between them, then `sin (theta/2)` is:

Options

A. `1/2 |vec(e)_(1)+vec(e)_(2)|`
B. `1/2 |vec(e)_(1)-vec(e)_(2)|`
C. `(vec(e)_(1).vec(e)_(2))/(2)`
D. `(|vec(e)_(1)xxvec(e)_(2)|)/(2|vec(e)_(1)||vec(e)_(2)|)`

Explanation

To find the value of \( \sin(\theta/2) \) where \( \theta \) is the angle between two unit vectors \( \vec{e_1} \) and \( \vec{e_2} \), we can use the following steps:

### Step 1: Understand the relationship between the angle and the dot product
The dot product of two unit vectors \( \vec{e_1} \) and \( \vec{e_2} \) is given by:
\[ \vec{e_1} \cdot \vec{e_2} = |\vec{e_1}| |\vec{e_2}| \cos(\theta) \]
Since both vectors are unit vectors, we have:
\[ \vec{e_1} \cdot \vec{e_2} = \cos(\theta) \]

### Step 2: Use the half-angle identity for sine
We can use the half-angle identity for sine, which states:
\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{2}} \]

### Step 3: Substitute the value of \( \cos(\theta) \)
From Step 1, we know that \( \cos(\theta) = \vec{e_1} \cdot \vec{e_2} \). Therefore, we can substitute this into the half-angle identity:
\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \vec{e_1} \cdot \vec{e_2}}{2}} \]

### Step 4: Final expression
Thus, the final expression for \( \sin\left(\frac{\theta}{2}\right) \) in terms of the dot product of the two unit vectors is:
\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \vec{e_1} \cdot \vec{e_2}}{2}} \]