Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular vectors of equal magnitudes, then the angle between the vectors $\overrightarrow{a}$ and $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. ${\cos }^{-1}\frac{1}{\sqrt{3}}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C ${\cos }^{-1}\frac{1}{\sqrt{3}}$

Since $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$ are mutually perpendicular, so $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$
Angle between $\overrightarrow{a}$ and $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$ is given by
$\cos \theta =\frac{\overrightarrow{a}.(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})}{|\overrightarrow{a}||\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}=\frac{a^2}{|\overrightarrow{a}||\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|}$
Now we know, $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=x$
$|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2+|\overrightarrow{c}{|}^2+2|\overrightarrow{a}|.|\overrightarrow{b}|+2|\overrightarrow{b}|.|\overrightarrow{c}|+2|\overrightarrow{c}|.|\overrightarrow{a}|$
$=x^2+x^2+x^2+0+0+0$
$\Rightarrow |\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}|=\sqrt{3}x$
$\Rightarrow \cos \theta =\frac{x^2}{x.\left(\sqrt{3}x\right)}$
$\Rightarrow \theta ={\cos }^{-1}\frac{1}{\sqrt{3}}$