Question

If $\overline{a},\overline{b}$ and $\overline{c}$ are perpendicular to $\overline{b}+\overline{c},\overline{c}+\overline{a}$ and $\overline{a}+\overline{b}$ respectively and If $|\overline{a}+\overline{b}|=6,|\overline{b}+\overline{c}|=8$ and $|\overline{c}+\overline{a}|=10$, then $|\overline{a}+\overline{b}+\overline{c}|$ is equal to

Answer 1

Given that,

$\begin{array}{c}\left|\overline{a}+\overline{b}\right|=6\left|\overline{b}+\overline{c}\right|=8\\ {\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+2\overline{a}.\overline{b}=36{\left|\overline{b}\right|}^2+{\left|\overline{c}\right|}^2+2\overline{b}\overline{c}=64\\ \left|\overline{c}+\overline{a}\right|=10\\ {\left|\overline{c}\right|}^2+{\left|\overline{a}\right|}^2+2\overline{a}\overline{c}=100\\ 2\left({\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+{\left|\overline{c}\right|}^2\right)+2\left(\overline{a}\overline{b}+\overline{b}\overline{c}+\overline{c}\overline{a}\right)=36+64+100\\ \\ \overline{a}.\left(\overline{b}+\overline{c}\right)=0\overline{a}.\overline{b}+\overline{a}.\overline{c}=0\\ \overline{b}.\left(\overline{c}+\overline{a}\right)=0\overline{b}.\overline{c}+\overline{b}.\overline{a}=0\\ \overline{c}.\left(\overline{a}+\overline{b}\right)=0\overline{c}.\overline{a}+\overline{c}.\overline{b}=0\\ 2\left(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a}\right)=0\\ \\ 2\left({\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+{\left|\overline{c}\right|}^2\right)=200\\ {\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+{\left|\overline{c}\right|}^2=100\\ {\left|\overline{a}+\overline{b}+\overline{c}\right|}^2=\left(\overline{a}+\overline{b}+\overline{c}\right).\left(\overline{a}+\overline{b}+\overline{c}\right)={\left|\overline{a}\right|}^2+{\left|\overline{b}\right|}^2+{\left|\overline{c}\right|}^2+2\left(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a}\right)=100\\ {\left|\overline{a}+\overline{b}+\overline{c}\right|}^2=100\end{array}$

Then,

We get $\left|\overline{a}+\overline{b}+\overline{c}\right|=10$