Question

If $\overline{a},\overline{b}$ and $\overline{a}+\overline{b}$ are unit vectors, then prove that the angle between $\overline{a}$ and $\overline{b}$ is $\frac{2\pi }{3}$.

Answer 1

Given : $\overline{a},\overline{b}$ and $\overline{a}+\overline{b}$ are unit vectors

$\therefore |\overline{a}|=1,|\overline{b}|=1$ and $|\overline{a}+\overline{b}|=1$

Since, $|\overline{a}+\overline{b}|=1$

$⟹|\overline{a}+\overline{b}{|}^2=1$

$⟹|\overline{a}{|}^2+|\overline{b}{|}^2+2|\overline{a}\cdot \overline{b}|=1$

$⟹1+1+2|\overline{a}\cdot \overline{b}|=1$

$⟹|\overline{a}\cdot \overline{b}|=-\frac{1}{2}$

We know that, the angle between two vectors $\overline{a}$ and $\overline{b}$ is given by

$\cos \theta =\frac{\overline{a}\cdot \overline{b}}{|\overline{a}||\overline{b}|}$

$=\frac{-1}{2\times 1\times 1}=-\frac{1}{2}$

$⟹-\cos \theta =\frac{1}{2}$

$⟹\cos (\pi -\theta )=\frac{1}{2}$

$⟹(\pi -\theta )={\cos }^{-1}\left(\frac{1}{2}\right)$

$⟹(\pi -\theta )=\frac{\pi }{3}$

$⟹\theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}$

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