If two unit vectors `vec(p) and vec(q)` make an angle `(pi)/(3)` with each other, what is the magnitude of `vec(p)-(1)/(2) vec(q)` ?

Options

A. `0`
B. `(sqrt(3))/(2)`
C. `1`
D. `(1)/(sqrt(2))`

Explanation

To find the magnitude of the vector \(\vec{p} - \frac{1}{2} \vec{q}\), where \(\vec{p}\) and \(\vec{q}\) are unit vectors making an angle of \(\frac{\pi}{3}\) with each other, we can follow these steps:

### Step 1: Understand the vectors
Since \(\vec{p}\) and \(\vec{q}\) are unit vectors, we have:
\[ |\vec{p}| = 1 \quad \text{and} \quad |\vec{q}| = 1 \]

### Step 2: Use the formula for the magnitude of a vector
We want to find the magnitude of \(\vec{p} - \frac{1}{2} \vec{q}\). The magnitude of a vector \(\vec{a}\) can be calculated using the formula:
\[ |\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}} \]
Thus, we need to compute:
\[ |\vec{p} - \frac{1}{2} \vec{q}| = \sqrt{(\vec{p} - \frac{1}{2} \vec{q}) \cdot (\vec{p} - \frac{1}{2} \vec{q})} \]

### Step 3: Expand the dot product
Expanding the dot product, we have:
\[ (\vec{p} - \frac{1}{2} \vec{q}) \cdot (\vec{p} - \frac{1}{2} \vec{q}) = \vec{p} \cdot \vec{p} - \vec{p} \cdot \left(\frac{1}{2} \vec{q}\right) - \left(\frac{1}{2} \vec{q}\right) \cdot \vec{p} + \left(\frac{1}{2} \vec{q}\right) \cdot \left(\frac{1}{2} \vec{q}\right) \]

### Step 4: Substitute known values
Using the properties of dot products:
- \(\vec{p} \cdot \vec{p} = |\vec{p}|^2 = 1\)
- \(\vec{q} \cdot \vec{q} = |\vec{q}|^2 = 1\)
- \(\vec{p} \cdot \vec{q} = |\vec{p}| |\vec{q}| \cos(\theta) = 1 \cdot 1 \cdot \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\)

Now substituting these values into the expanded dot product:
\[ = 1 - \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4} \]
\[ = 1 - \frac{1}{4} - \frac{1}{4} + \frac{1}{4} \]
\[ = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 5: Take the square root
Now we take the square root to find the magnitude:
\[ |\vec{p} - \frac{1}{2} \vec{q}| = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Final Answer
Thus, the magnitude of \(\vec{p} - \frac{1}{2} \vec{q}\) is:
\[ \frac{\sqrt{3}}{2} \]