If x and y are two unit vectors and `phi` is the angle between them, then `(1)/(2) |x-y|` is equal to

Options

A. 0
B. `|sin(phi/2)|`
C. `|sin(1)/(2)phi|`
D. `|cos(1)/(2)phi|`

Answer

To solve the problem, we need to find the expression for \(\frac{1}{2} |x - y|\) where \(x\) and \(y\) are unit vectors and \(\phi\) is the angle between them.

### Step-by-Step Solution:

1. **Understanding the Magnitude of the Difference of Two Vectors**:
We start with the expression \(|x - y|\). The magnitude of the difference of two vectors can be expressed using the formula:
\[ |x - y|^2 = |x|^2 + |y|^2 - 2(x \cdot y) \]

2. **Substituting the Values**:
Since \(x\) and \(y\) are unit vectors, we have:
\[ |x|^2 = 1 \quad \text{and} \quad |y|^2 = 1 \]
Therefore, substituting these values into the equation gives:
\[ |x - y|^2 = 1 + 1 - 2(x \cdot y) = 2 - 2(x \cdot y) \]

3. **Using the Dot Product**:
The dot product \(x \cdot y\) can be expressed in terms of the angle \(\phi\) between them:
\[ x \cdot y = |x||y|\cos(\phi) = 1 \cdot 1 \cdot \cos(\phi) = \cos(\phi) \]
Substituting this into our equation gives:
\[ |x - y|^2 = 2 - 2\cos(\phi) \]

4. **Simplifying the Expression**:
We can factor out the 2:
\[ |x - y|^2 = 2(1 - \cos(\phi)) \]

5. **Taking the Square Root**:
To find \(|x - y|\), we take the square root of both sides:
\[ |x - y| = \sqrt{2(1 - \cos(\phi))} \]

6. **Using the Trigonometric Identity**:
We can use the trigonometric identity \(1 - \cos(\phi) = 2\sin^2\left(\frac{\phi}{2}\right)\):
\[ |x - y| = \sqrt{2 \cdot 2\sin^2\left(\frac{\phi}{2}\right)} = \sqrt{4\sin^2\left(\frac{\phi}{2}\right)} = 2\sin\left(\frac{\phi}{2}\right) \]

7. **Finding \(\frac{1}{2} |x - y|\)**:
Finally, we multiply by \(\frac{1}{2}\):
\[ \frac{1}{2} |x - y| = \frac{1}{2} \cdot 2\sin\left(\frac{\phi}{2}\right) = \sin\left(\frac{\phi}{2}\right) \]

### Final Answer:
Thus, the value of \(\frac{1}{2} |x - y|\) is:
\[ \sin\left(\frac{\phi}{2}\right) \]