Question
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three mutually perpendicular unit vectors, then prove that $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}$

Answer

Given that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are unit vectors.
So, $\big|\vec{\text{a}}\big|=1,\Big|\vec{\text{b}}\big|=1$ and $\big|\vec{\text{c}}\big|=1$
Since they are mutually perpendicular,
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|+2\vec{\text{a}}.\vec{\text{b}}+2\vec{{\text{b}}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}$
$=1+1+1+0+0+0$
$=3$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}$

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