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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA3 Marks
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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