# If hat(a_1),hat(a_2)and hat(a_3) are unit vectors and 2hat(a_1)+2hat(a_2)+hat(a_3) =0 then angle between hat(a_1) and hat(a_2) is ?

Jun 10, 2018

Given $\hat{{a}_{1}} , \hat{{a}_{2}} \mathmr{and} \hat{{a}_{3}}$ are unit vectors

So $\left\mid \hat{{a}_{1}} \right\mid = \left\mid \hat{{a}_{2}} \right\mid = \left\mid \hat{{a}_{3}} \right\mid = 1$

Let the angle between $\hat{{a}_{1}} \mathmr{and} \hat{{a}_{2}}$ is $\theta$

Again given

$2 \hat{{a}_{1}} + 2 \hat{{a}_{2}} + \hat{{a}_{3}} = 0. \ldots . \left[1\right]$

So $2 \left\mid \hat{{a}_{1}} + \hat{{a}_{2}} \right\mid = \left\mid \hat{{a}_{3}} \right\mid$

$\implies 2 \sqrt{{\left\mid \hat{{a}_{1}} \right\mid}^{2} + {\left\mid \hat{{a}_{2}} \right\mid}^{2} + 2 \left\mid \hat{{a}_{1}} \right\mid \left\mid \hat{{a}_{2}} \right\mid \cos \theta} = \left\mid \hat{{a}_{3}} \right\mid$

$\implies 2 \sqrt{{1}^{2} + {1}^{2} + 2 \cdot 1 \cdot 1 \cdot \cos \theta} = 1$

$\implies 2 \sqrt{2 + 2 \cos \theta} = 1$

$\implies 8 \left(1 + \cos \theta\right) = 1$

$\implies \cos \theta = \frac{1}{8} - 1 = - \frac{7}{8}$

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

Alternative way

We have

$2 \hat{{a}_{1}} + 2 \hat{{a}_{2}} + \hat{{a}_{3}} = 0$

$\implies 2 \hat{{a}_{1}} \cdot \hat{{a}_{1}} + 2 \hat{{a}_{2}} \cdot \hat{{a}_{1}} + \hat{{a}_{3}} \hat{{a}_{1}} = 0$

$\implies 2 + 2 \left\mid \hat{{a}_{2}} \right\mid \left\mid \hat{{a}_{1}} \right\mid \cos \theta + \hat{{a}_{3}} \hat{{a}_{1}} = 0$

$\textcolor{red}{\implies 2 + 2 \cos \theta + \hat{{a}_{3}} \hat{{a}_{1}} = 0. \ldots \ldots \left[2\right]}$

similarly

$2 \hat{{a}_{1}} \cdot \hat{{a}_{2}} + 2 \hat{{a}_{2}} \cdot \hat{{a}_{2}} + \hat{{a}_{3}} \cdot \hat{{a}_{2}} = 0$

$\textcolor{b l u e}{\implies 2 \cos \theta + 2 + \hat{{a}_{3}} \cdot \hat{{a}_{2}} = 0. \ldots . . \left[3\right]}$

And also

$2 \hat{{a}_{1}} \cdot \hat{{a}_{3}} + 2 \hat{{a}_{2}} \cdot \hat{{a}_{3}} + \hat{{a}_{3}} \cdot \hat{{a}_{3}} = 0$

=>color(green)(2hat(a_1)*hat(a_3)+2hat(a_2)*hat(a_3)+1)=0.......[4])

Combining [2],[3]and [4] we get

$8 \left(1 + \cos \theta\right) = 1$

$\implies \cos \theta = \frac{1}{8} - 1 = - \frac{7}{8}$

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

Shortest method

$2 \hat{{a}_{1}} + 2 \hat{{a}_{2}} = - \hat{{a}_{3}}$

$\implies \left(2 \hat{{a}_{1}} + 2 \hat{{a}_{2}}\right) \cdot \left(2 \hat{{a}_{1}} + 2 \hat{{a}_{2}}\right) = \hat{{a}_{3}} \cdot \hat{{a}_{3}}$

$\implies 4 \left(1 + 2 \hat{{a}_{2}} \cdot \hat{{a}_{1}} + 1\right) = 1$

$\implies 8 \left(1 + \cos \theta\right) = 1$

$\implies \cos \theta = \frac{1}{8} - 1 = - \frac{7}{8}$

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

Jun 10, 2018

$\boldsymbol{\phi = {\cos}^{- 1} \left(- \frac{7}{8}\right)}$

#### Explanation:

$\hat{{a}_{1}} \cdot \hat{{a}_{2}} = \cos \phi$

$2 \hat{{a}_{1}} + 2 \hat{{a}_{2}} + \hat{{a}_{3}} = 0 \implies \hat{{a}_{1}} = - \hat{{a}_{2}} - \frac{1}{2} \hat{{a}_{3}}$

Using: $q \quad \boldsymbol{\hat{{a}_{1}} = - \hat{{a}_{2}} - \frac{1}{2} \hat{{a}_{3}}}$

• $\boldsymbol{\hat{{a}_{1}} \cdot \hat{{a}_{1}}} = 1 = - \hat{{a}_{1}} \cdot \hat{{a}_{2}} - \frac{1}{2} \hat{{a}_{1}} \cdot \hat{{a}_{3}} = - \cos \phi - \frac{1}{2} \hat{{a}_{1}} \cdot \hat{{a}_{3}}$

$\implies \cos \phi = - 1 - \frac{1}{2} \hat{{a}_{1}} \cdot \hat{{a}_{3}} q \quad \square$

• $\boldsymbol{\hat{{a}_{1}} \cdot \hat{{a}_{2}}} = - {a}_{2}^{2} - \frac{1}{2} \hat{{a}_{2}} \cdot \hat{{a}_{3}} = - 1 - \frac{1}{2} \hat{{a}_{2}} \cdot \hat{{a}_{3}}$

$\implies \cos \phi = - 1 - \frac{1}{2} \hat{{a}_{2}} \cdot \hat{{a}_{3}} q \quad \triangle$

• $\boldsymbol{\hat{{a}_{1}} \cdot \hat{{a}_{3}}} = - \hat{{a}_{2}} \cdot \hat{{a}_{3}} - \frac{1}{2} q \quad \circ$

$\square + \circ \implies \cos \phi = - 1 - \frac{1}{2} \left(- \hat{{a}_{2}} \cdot \hat{{a}_{3}} - \frac{1}{2}\right)$

$= - \frac{3}{4} + \frac{1}{2} \hat{{a}_{2}} \cdot \hat{{a}_{3}} q \quad \star$

$\triangle + \star \implies 2 \cos \phi = - \frac{7}{4}$

$\boldsymbol{\phi = {\cos}^{- 1} \left(- \frac{7}{8}\right)}$