Question

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

• A. $\sqrt{3}$
• B. $3$
• C. $1$
• D. $0$

Answer 1

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

Since, $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular unit vectors.
$\Rightarrow \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1$
and $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{c}\cdot \overrightarrow{a}=0$
Now,
${|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}^2=(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\cdot (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$
$={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2+2(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a})$
$=1+1+1+0=3$
$\therefore |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{3}$