Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three mutually perpendicular unit vectors and $d$ is a unit vector making equal angles with $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},$ then ${|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}|}^2$ is

• A. $4$
• B. $4\pm \sqrt{3}$
• C. $4\pm 2\sqrt{3}$
• D. None of these

Answer 1

Answer: (C)

$4\pm 2\sqrt{3}$

We have

$|\overline{a}+\overline{b}+\overline{c}+\overline{d}{|}^2=4+2[\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{d}.+\overline{d}.\overline{a}+\overline{a}.\overline{c}+\overline{b}.\overline{d}]$
$=4+2(0+0+0+\overline{d}(\overline{a}+\overline{b}+\overline{c}\left)\right)$
Let $\overline{d}=\lambda \overline{a}+\mu \overline{b}+v\overline{c}$
$\overline{d}.\overline{a}=\lambda ,\overline{d}.\overline{b}=\mu ,\overline{d}.\overline{c}=v$
But $\overline{d}.\overline{a}=\overline{d}.\overline{b}=\overline{d}.\overline{c}=\cos \theta$
But ${\lambda }^2+{\mu }^2+v^2=1$
$3{\cos }^2\theta =1$

$\Rightarrow cos\theta =(\pm \frac{1}{\sqrt{3}})$
$|\overline{a}+\overline{b}+\overline{c}+\overline{d}{|}^2=4+2$. $(\pm \frac{1}{\sqrt{3}})$
$=4\pm 2\sqrt{3}$