Question

Let $\vec{A},\vec{B},\vec{C}$ be vectors of length $3,4,5$ respectively. Let $\vec{A}$ be perpendicular to $\vec{B}+\vec{C},\vec{B}$ to $\vec{C}+\vec{A}$ and $\vec{C}$ to $\vec{A}+\vec{B}$ Then, the length of vector$\vec{A}+\vec{B}+\vec{C}$ is....

Answer 1

$|\overrightarrow{a}|=3,|\overrightarrow{d}|=4,|\overrightarrow{b}|=5$
Since, $\overrightarrow{A}.(\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{B}(\overrightarrow{C}+\overrightarrow{A})=\overrightarrow{C}.(\overrightarrow{A}+\overrightarrow{B})=\mathrm{0...}(i)$
$\therefore |\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}{|}^2=|\overrightarrow{A}|+|\overrightarrow{B}{|}^2+|\overrightarrow{C}{|}^2+(\overrightarrow{A}.\overrightarrow{B}+\overrightarrow{B}.\overrightarrow{C}+\overrightarrow{C}.\overrightarrow{A})$
$=9+16+25+0$
[ from Eq.$(i)\overrightarrow{A}.\overrightarrow{B}+\overrightarrow{B}.\overrightarrow{C}+\overrightarrow{C}.\overrightarrow{A}=0]$
$\therefore |\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}{|}^2=50$
$\Rightarrow |\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}|=5\sqrt{2}$