Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$are any three mutually perpendicular vectors of equal magnitude a, then $\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|$ is equal to
(a) a
(b) $\sqrt{2}a$
(c) $\sqrt{3}a$
(d) 2a
(e) None of these

Answer 1

(c) $\sqrt{3}a$

$$\begin{gathered} \text{Given that} \\ \text{So,}\left|\overrightarrow{a}\right|\text{=}\left|\overrightarrow{b}\right|\text{=}\left|\overrightarrow{c}\right|\text{=}a...\left(\mathrm{i}\right) \\ \text{Since they are mutually perpendicular,} \\ \overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0...\left(\mathrm{ii}\right) \\ \text{Now,} \\ {\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}^2={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}+2\overrightarrow{b}.\overrightarrow{c}+2\overrightarrow{c}.\overrightarrow{a} \\ =a^2+a^2+a^2+0+0+0\left[\mathrm{Using}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\right] \\ =3a^2 \\ \therefore \left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=\sqrt{3}a \end{gathered}$$