Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three mutually perpendicular vectors of equal magnitude, then the angle $\theta$ which $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ makes with any one of three given vectors is given by

• A. $co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• B. $co{s}^{-1}\left(\frac{1}{3}\right)$
• C. $co{s}^{-1}\left(\frac{2}{\sqrt{3}}\right)$
• D. $Noneofthese$

Answer 1

The correct option is A $co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

It is given that, $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=\lambda$ (say) and $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular vectors.
Therefore,$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{3}\lambda$
Let $\theta$ be the angle which $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ makes with $\overrightarrow{a}$.
Then,
$cos\theta =\frac{\overrightarrow{a}.(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})}{|\overrightarrow{a}||\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}=\frac{|\overrightarrow{a}{|}^2}{|\overrightarrow{a}||\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}=\frac{{\lambda }^2}{\lambda \left(\sqrt{3}\lambda \right)}=\frac{1}{\sqrt{3}}\therefore \theta =co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$