# If hata_1,hata_2andhata_3 are unit vectors and2hata_1+2hata_2+hata_3=0 then angle between hata_1 and hata_2  is?

Jun 23, 2018

${\cos}^{-} 1 \left(- \frac{7}{8}\right) \approx {151}^{\circ}$

#### Explanation:

${\hat{a}}_{3} = - 2 \left({\hat{a}}_{1} + {\hat{a}}_{2}\right) \implies$
${\hat{a}}_{3}^{2} = 4 {\left({\hat{a}}_{1} + {\hat{a}}_{2}\right)}^{2}$
$q \quad = 4 \left({\hat{a}}_{1}^{2} + {\hat{a}}_{2}^{2} + 2 {\hat{a}}_{1} \cdot {\hat{a}}_{2}\right) \implies$
$1 = 4 \left(1 + 1 + 2 {\hat{a}}_{1} \cdot {\hat{a}}_{2}\right) \implies$
${\hat{a}}_{1} \cdot {\hat{a}}_{2} = - \frac{7}{8}$

Thus, the angle $\theta$ between ${\hat{a}}_{1}$ and ${\hat{a}}_{2}$ is given by

$\cos \theta = \frac{{\hat{a}}_{1} \cdot {\hat{a}}_{2}}{| {\hat{a}}_{1} | \setminus | {\hat{a}}_{2} |} = - \frac{7}{8}$ and thus

$\theta = {\cos}^{-} 1 \left(- \frac{7}{8}\right) \approx {151}^{\circ}$