Question

Let $\overline{a},\overline{b},\overline{c}$ be vectors of length 3,4,5 respectively. Let $\overline{a}$ be perpendicular to $\overline{b}+\overline{c}$,$\overline{b}$ is perpendicular to $\overline{c}+\overline{a}$ & $\overline{c}$ is perpendicular to $\overline{a}+\overline{b}$. Then $|\overline{a}+\overline{b}+\overline{c}|$ is:

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

Answer 1

The correct option is D $5\sqrt{2}$

Given,

$|\overrightarrow{a}|=3,|\overrightarrow{b}|=4,|\overrightarrow{c}|=5$

And

$\overrightarrow{a}.(\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{b}.(\overrightarrow{c}+\overrightarrow{a})=\overrightarrow{c}.(\overrightarrow{a}+\overrightarrow{b})$

Now,

$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$

$=\overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.(\overrightarrow{b}+\overrightarrow{c})+\overrightarrow{b}.\overrightarrow{b}+\overrightarrow{b}.(\overrightarrow{a}+\overrightarrow{c})+\overrightarrow{c}.\overrightarrow{c}+\overrightarrow{c}.(\overrightarrow{a}+\overrightarrow{b})$

$=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2$

$=9+16+25=50$

$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{50}=5\sqrt{2}$

Hence, the correct answer is $D$.