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Q.
If $|\bar{a}|=3,|\bar{b}|=4,|\bar{c}|=5$ and every vector $\bar{a}$ is perpendicular to the sum of other two vectors, then find the value of $|\bar{a}+\bar{b}+\bar{c}|$.
Vector Algebra
Solution:
$|\bar{a}|=3,|\bar{b}|=4$ and $|\bar{c}|=5 $... [Given]
Every vector is perpendicular to the sum of other two vectors.
$\Rightarrow \bar{a} \cdot(\bar{b}+\bar{c})=0, \bar{b} \cdot(\bar{c}+\bar{a})=0 $ and $\bar{c} \cdot(\bar{a}+\bar{b})=0$
$ \Leftrightarrow \bar{a} \cdot \bar{b}+\bar{a} \cdot \bar{c}=0, \bar{b} \cdot \bar{c}+\bar{b} \cdot \bar{a}= 0 $ and$ \bar{c} \cdot \bar{a}+\bar{c} \cdot \bar{b}=0$
Adding all three,
$\Rightarrow 2(\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a})=0\,\,\,(i)$
We know that,
$|\bar{a}+\bar{b}+\bar{c}|^{2} =(\bar{a}+\bar{b}+\bar{c}) \cdot(\bar{a}+\bar{b}+\bar{c})$
$=|\bar{a}|^{2}+|\bar{b}|^{2}+|\bar{c}|^{2} +2(\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}) $
$=9+16+25+0\ldots[$ From( i )$]$
$\Rightarrow|\bar{a}+\bar{b}+\bar{c}|^{2}=50$