Question

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

Answer 1

Given:
$\overrightarrow{a},\overrightarrow{b}\mathrm{and}\overrightarrow{c}$ are unit vectors
$\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{0}$


$$\begin{gathered} \overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{0} \\ \Rightarrow \overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c} \\ \Rightarrow {\left(\overrightarrow{a}+\overrightarrow{b}\right)}^2={\left(\overrightarrow{c}\right)}^2 \\ \Rightarrow \left(\overrightarrow{a}+\overrightarrow{b}\right)·\left(\overrightarrow{a}+\overrightarrow{b}\right)=1\left(\because \left|\overrightarrow{c}\right|=1\right) \\ \Rightarrow \overrightarrow{a}·\overrightarrow{a}+\overrightarrow{b}·\overrightarrow{b}+2\overrightarrow{a}·\overrightarrow{b}=1 \\ \Rightarrow {\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left(\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \right)=1 \\ \Rightarrow 1^2+1^2+2\left(\left(1\right)\left(1\right)\cos \theta \right)=1\left(\because \left|\overrightarrow{a}\right|=1,\left|\overrightarrow{b}\right|=1\right) \\ \Rightarrow 1+1+2\cos \theta =1 \\ \Rightarrow 2\cos \theta =-1 \\ \Rightarrow \cos \theta =-\frac{1}{2} \\ \Rightarrow \theta =\mathrm{\pi }-\frac{\mathrm{\pi }}{3} \\ \Rightarrow \theta =\frac{2\mathrm{\pi }}{3} \end{gathered}$$

Hence, the angle between $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ is $\underline{\frac{2\mathrm{\pi }}{3}}$.