Question

If for non-zero vectors $\overline{a},\overline{b},\overline{c}$ $\overline{a}\times \overline{b}=\overline{c}$ and $\overline{b}\times \overline{c}=\overline{a}$, then prove that $|\overline{b}|=1$.

Answer 1

Here $\overline{b}\times \overline{c}=\overline{a}$
$\therefore (\overline{b}\times \overline{c})\cdot \overline{b}=\overline{a}\cdot \overline{b}$
$\therefore \left[\overline{b}\overline{c}\overline{b}\right]=\overline{a}\cdot \overline{b}$
($\therefore$ Definition of Tripple product)
$\therefore \overline{a}\cdot \overline{b}=0$ ..$\left(1\right)$
And $\overline{b}\times \overline{c}=\overline{a}$
$\therefore \overline{b}\times (\overline{a}\times \overline{b})=\overline{a}$
$(\because \overline{c}=\overline{a}\times \overline{b})$
$\therefore (\overline{b}\cdot \overline{b})\overline{a}-(\overline{b}-\overline{b})\overline{b}=\overline{a}$
$\therefore |\overline{b}{|}^2\cdot \overline{a}=\overline{a}$
$\therefore (|\overline{b}{|}^2\cdot \overline{a}-\overline{a})=\overline{0}$
But $\overline{a}\ne \overline{0}$
$\therefore \overline{a}(|\overline{b}{|}^2-1)=\overline{0}$
$\therefore |\overline{b}{|}^2-1=0$
$\therefore |\overline{b}{|}^2=1$
$\therefore |\overline{b}=1$