Question

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

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

Answer 1

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

Solution:

Given that: $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular .i.e

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

Also, $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$

Solution:

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

$=\sqrt{1+1+1+0+0+0}$

$=\sqrt{3}$

Hence, B is the correct option.