Mathematics - Vector Question with Solution | TestHub

Given, 

$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$

$\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}$

$\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{b}$

$(\overrightarrow{b}\times \overrightarrow{c})·\overrightarrow{a}=\overrightarrow{a}·\overrightarrow{a}$

$(\overrightarrow{c}\times \overrightarrow{a})·\overrightarrow{b}=\overrightarrow{b}·\overrightarrow{b}$

${\overrightarrow{a}}^2-{\overrightarrow{b}}^2=(\overrightarrow{b}\times \overrightarrow{c})·\overrightarrow{a}-(\overrightarrow{c}\times \overrightarrow{a})·\overrightarrow{b}$

$=\left[\begin{array}{ccc}\overrightarrow{c}& \overrightarrow{a}& \overrightarrow{b}\end{array}\right]-\left[\begin{array}{ccc}\overrightarrow{b}& \overrightarrow{c}& \overrightarrow{a}\end{array}\right]=0$

$\therefore |\overrightarrow{a}{|}^2-|\overrightarrow{b}{|}^2=0$

$\left|\overrightarrow{a}\right|-\left|\overrightarrow{b}\right|=0$

$\Rightarrow \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|$

We know that, 

$\left[\overrightarrow{a}\times \overrightarrow{b} \overrightarrow{b}\times \overrightarrow{c} \overrightarrow{c}\times \overrightarrow{a}\right]=[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}{]}^2$

$\left[\begin{array}{l}\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\end{array}\right]={\left[\begin{array}{lll}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]}^2$

$\Rightarrow \left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]=1$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|=1$

$\therefore \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|\ne 2$