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Q. $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ and $|\overrightarrow{a}|=5,|\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=3$ then the value of $|\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}|$ is

Vector Algebra

Solution:

Given, $\vec{a}+\vec{b}+\vec{c}=\vec{0}$
$\therefore \left|\vec{a}\right|^{2}+\left|\vec{b}\right|^{2}+\left|\vec{c}\right|^{2}-2\,\vec{a}\cdot\vec{c}-2\,\vec{b}\cdot\vec{c}+2\,\vec{a}\cdot\vec{c}$
$=25+16+9+2\left[\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}-\vec{a}\cdot\vec{c}\right]=0$
$\Rightarrow 2\left[\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}-\vec{c}\cdot\vec{a}\right]=-50$
$\Rightarrow \left[\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}-\vec{c}\cdot\vec{a}\right]=-25$