If a + b + c =0, show that the angle between the vectors b and c is given by = | a |^2- | b |^2-| c |^2 2 | b || c | .

Given, a + b + c =0 b + c =- a | b + c |^2=|- a |^2 | b |^2+| c |^2+2 b . c =| a |^2 2 b . c =| a |^2- | b |^2-| c |^2 2 | b || c | =| a |^2- | b |^2-| c |^2 = | a |^2- | b |^2-| c |^2 2 | b || c |

If a + b + c =0, show that the angle between the vectors b and c …

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Question

If $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{0,}$ show that the angle $\theta$ between the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$ is given by $\cos\theta=\frac{|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2-|\vec{\text{c}}|^2}{2\big|\vec{\text{b}}\big||\vec{\text{c}}|}.$

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Answer

Given,$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$
$\Rightarrow\vec{\text{b}}+\vec{\text{c}}=-\vec{\text{a}}$
$\Rightarrow\big|\vec{\text{b}}+\vec{\text{c}}\big|^2=|-\vec{\text{a}}|^2$
$\Rightarrow\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|^2+2\vec{\text{b}}.\vec{{\text{c}}}=|\vec{\text{a}}|^2$
$\Rightarrow2\vec{\text{b}}.\vec{\text{c}}=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2-|\vec{\text{c}}|^2$
$\Rightarrow2\big|\vec{\text{b}}\big||\vec{\text{c}}|\cos\theta=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2-|\vec{\text{c}}|^2$
$\therefore\cos\theta=\frac{|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2-|\vec{\text{c}}|^2}{2\big|\vec{\text{b}}\big||\vec{\text{c}}|}$

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