The angle (in degree) between two planes vectors \[\overrightarrow {\rm{a}} = \frac{{\sqrt 3 }}{2}{\rm{\hat i}} + \frac{1}{2}{\rm{\hat j}}\] and \[\overrightarrow {\rm{b}} = \frac{{ – \sqrt 3 }}{2}{\rm{\hat i}} + \frac{1}{2}{\rm{\hat j}}\] is A. 30 B. 60 C. 90 D. 120

30
60
90
120

The correct answer is $\boxed{\text{C}}$.

The angle between two vectors $\overrightarrow{\rm{a}}$ and $\overrightarrow{\rm{b}}$ is given by the following formula:

$$\cos \theta = \frac{\overrightarrow{\rm{a}} \cdot \overrightarrow{\rm{b}}}{\left\| \overrightarrow{\rm{a}} \right\| \left\| \overrightarrow{\rm{b}} \right\|}$$

where $\theta$ is the angle between the two vectors, $\overrightarrow{\rm{a}} \cdot \overrightarrow{\rm{b}}$ is the dot product of the two vectors, and $\left\| \overrightarrow{\rm{a}} \right\|$ and $\left\| \overrightarrow{\rm{b}} \right\|$ are the norms of the two vectors.

In this case, we have:

$$\overrightarrow{\rm{a}} = \frac{{\sqrt 3 }}{2}{\rm{\hat i}} + \frac{1}{2}{\rm{\hat j}}$$

$$\overrightarrow{\rm{b}} = \frac{{ – \sqrt 3 }}{2}{\rm{\hat i}} + \frac{1}{2}{\rm{\hat j}}$$

Therefore, we have:

$$\overrightarrow{\rm{a}} \cdot \overrightarrow{\rm{b}} = \frac{{\sqrt 3 }}{2} \cdot \frac{{ – \sqrt 3 }}{2} + \frac{1}{2} \cdot \frac{1}{2} = -\frac{3}{4}$$

$$\left\| \overrightarrow{\rm{a}} \right\| = \sqrt{\left(\frac{{\sqrt 3 }}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}$$

$$\left\| \overrightarrow{\rm{b}} \right\| = \sqrt{\left(\frac{{ – \sqrt 3 }}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}$$

Therefore, we have:

$$\cos \theta = \frac{\overrightarrow{\rm{a}} \cdot \overrightarrow{\rm{b}}}{\left\| \overrightarrow{\rm{a}} \right\| \left\| \overrightarrow{\rm{b}} \right\|} = \frac{-\frac{3}{4}}{\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}} = -\frac{3}{3} = -1$$

The angle between two vectors is $\boxed{90^\circ}$ when the dot product of the two vectors is equal to $-1$.