If \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) are unit vectors such that \(\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}\), then angle between \(\mathbf{a}\) and \(\mathbf{b}\) is

Given, \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are unit vectors.
\[
\begin{array}{lc}
\Rightarrow & |a|=|b|=|c|=1 \\
\text { Also, given } & \mathbf{a}+\mathbf{b}+\mathbf{c}=0 \\
\Rightarrow & (a+b)=-c
\end{array}
\]
Squaring on both sides, we get
\[
\begin{array}{lc}
\Rightarrow & (\mathbf{a}+\mathbf{b})^{2}=(\mathbf{c})^{2} \\
\Rightarrow & (\mathbf{a})^{2}+(\mathbf{b})^{2}+2 \mathbf{a} \cdot \mathbf{b}=(\mathbf{c})^{2} \\
\Rightarrow & |\mathbf{a}|^{2}+|\mathbf{c}|^{2}+2 \mathbf{a} \cdot \mathbf{b}=|\mathbf{c}|^{2} \\
& \quad\left[\because(\mathbf{a})^{2}=|\mathbf{a}|^{2}\right] \\
\Rightarrow & 1+1+2 \mathbf{a} \cdot \mathbf{b}=1[\text { from Eq. (i) }] \\
\Rightarrow & 2 \mathbf{b}=-1 \\
\Rightarrow & \mathbf{a} \cdot \mathbf{b}=-1 / 2=|\mathbf{a}||\mathbf{b}| \cos \theta \\
\Rightarrow & \cos 0=-1 / 2=\cos 2 \pi / 3 \quad[\text { from Eq. (i) }] \\
\Rightarrow & \theta=2 \pi / 3
\end{array}
\]