Question

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 $|\bar{a}+\bar{b}+\bar{c}|$.

Answer 1

Answer: 50

$|\bar{a}|=3,|\bar{b}|=4$ 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 $\bar{c}\cdot (\bar{a}+\bar{b})=0$
$\iff \bar{a}\cdot \bar{b}+\bar{a}\cdot \bar{c}=0,\bar{b}\cdot \bar{c}+\bar{b}\cdot \bar{a}=0$ 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,
$|\bar{a}+\bar{b}+\bar{c}{|}^2=(\bar{a}+\bar{b}+\bar{c})\cdot (\bar{a}+\bar{b}+\bar{c})$
$=|\bar{a}{|}^2+|\bar{b}{|}^2+|\bar{c}{|}^2+2(\bar{a}\cdot \bar{b}+\bar{b}\cdot \bar{c}+\bar{c}\cdot \bar{a})$
$=9+16+25+0\dots [$ From( i )$]$
$\Rightarrow |\bar{a}+\bar{b}+\bar{c}{|}^2=50$