Question

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

• A. $5\sqrt{2}$
• B. $50$
• C. $10\sqrt{2}$
• D. $10$

Answer 1

Answer: (D)

$10$

We have,
$\overline{a}.\overline{b}+\overline{a}.\overline{c}=0$ , $\overline{b}.\overline{c}+\overline{a}.\overline{b}=0$ and $\overline{a}.\overline{c}+\overline{b}.\overline{c}=0$
$\Rightarrow \overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a}=0$
$|a+b|=6$
$\Rightarrow |a{|}^2+|b{|}^2+2\overline{a}.\overline{b}=36$
Similarly, we will get two more equations and on adding the 3 equations we get,
$a^2+b^2+c^2+(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})=100$
$\Rightarrow |a{|}^2+|b{|}^2+|c{|}^2=100$
Now,
$|\overline{a}+\overline{b}+\overline{c}{|}^2$
$=|a{|}^2+|b{|}^2+|c{|}^2+2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{a}.\overline{c})=100$
$\Rightarrow |a+b+c{|}^2=100$
$\Rightarrow |a+b+c|=10$