Step 1: Understanding the Concept:
The angle \(\theta\) between two non-zero vectors \(\vec{a}\) and \(\vec{b}\) can be found using the definition of the dot product: \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\). We can rearrange this formula to solve for \(\cos\theta\).
Step 2: Key Formula or Approach:
The formula for the angle between two vectors is:
\[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]
We need to calculate three things: the dot product \(\vec{a} \cdot \vec{b}\), the magnitude \(|\vec{a}|\), and the magnitude \(|\vec{b}|\).
Step 3: Detailed Explanation:
Given vectors:
\[ \vec{a} = \frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j} \]
\[ \vec{b} = -\frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j} \]
1. Calculate the magnitudes:
\[ |\vec{a}| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{\frac{4}{4}} = \sqrt{1} = 1 \]
\[ |\vec{b}| = \sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{\frac{4}{4}} = \sqrt{1} = 1 \]
Both are unit vectors.
2. Calculate the dot product:
\[ \vec{a} \cdot \vec{b} = \left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \]
\[ \vec{a} \cdot \vec{b} = -\frac{3}{4} + \frac{1}{4} = -\frac{2}{4} = -\frac{1}{2} \]
3. Find the angle:
\[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-1/2}{(1)(1)} = -\frac{1}{2} \]
We need to find the angle \(\theta\) (between \(0^\circ\) and \(180^\circ\)) for which \(\cos\theta = -1/2\).
From trigonometry, we know that \(\cos(180^\circ - 60^\circ) = -\cos(60^\circ) = -1/2\).
So, \(\theta = 180^\circ - 60^\circ = 120^\circ\).
Step 4: Final Answer:
The angle between the two vectors is 120 degrees. This corresponds to option (D).