If $\displaystyle \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\displaystyle \overrightarrow{\mathrm{c}}$ are three unit vectors such that $\displaystyle \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=\overrightarrow{0}$, find the angle between vectors $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{c}}$.

Given $\displaystyle |\vec{\mathrm{a}}|=|\vec{\mathrm{b}}|=|\vec{\mathrm{c}}|=1$ Now $\displaystyle \vec{\mathrm{a}}-\vec{\mathrm{c}}=-\vec{\mathrm{b}}$ $\displaystyle (\vec{\mathrm{a}}-\vec{\mathrm{c}}) \cdot(\vec{\mathrm{a}}-\vec{\mathrm{c}})=(-\vec{\mathrm{b}}) \cdot(-\vec{\mathrm{b}})$ $\displaystyle \Rightarrow|\vec{\mathrm{a}}|^{2}+|\vec{\mathrm{c}}|^{2}-2 \vec{\mathrm{a}} \cdot \vec{\mathrm{c}}=|\vec{\mathrm{b}}|^{2}$ $\displaystyle \Rightarrow 1+1-2|\vec{\mathrm{a}}||\vec{\mathrm{c}}| \cos \theta=1$ $\displaystyle \Rightarrow 2-2(1)(1) \cos \theta=1$ $\displaystyle \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$