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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA5 Marks
Question
If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},$ are three vectors such that $|\overrightarrow{a}|=5,|\overrightarrow{b}|=12\text{ and }|\overrightarrow{c}|=13,$ and $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{o},$ find the value of $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}.$
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Answer

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\Rightarrow\Bigg(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\Bigg)^{2}=0$$\Rightarrow\overrightarrow{a}^{2}+\overrightarrow{b}^{2}+\overrightarrow{c}^{2}+2\Bigg(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\Bigg)=0$
OR
$|\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}+2\Bigg(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\Bigg)=0$
$\therefore\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=-\frac{1}{2}\Bigg(|\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}\Bigg)$
$=-\frac{1}{2}(25+144+169)=-169$.

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