Question

If $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are mutually perpendicular unit vectors, then $|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|$ is equal to :

• A. $3$
• B. $\sqrt{3}$
• C. $\frac{\left(\sqrt{a^2+b^2+c^2}\right)}{3}$
• D. $1$

Answer 1

The correct option is B $\sqrt{3}$

Given that:

$\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are mutually perpendicular unit vector.

So, $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$ and $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$

Now,

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

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

$=1^2+1^2+1^2+2(0+0+0)$

$=1+1+1+0$

$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=3$

$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{3}$

Hence, B is the correct option.