Hello every one I have this vector relation
E=A+(HXA) where E,A and H are vectors I want to proof that
A=(E-(HXA))/
any one remeber how we can do this using vector identities or simple geometry or both of them (may be)?
If that is the case then clearly $\displaystyle A~\& ~H$ are also perpendicular. Form which we get using a well known fact
$\displaystyle H \times \left( {H \times A} \right) = \left( {H \cdot A} \right)H - \left( {H \cdot H} \right)A = - \left\| H \right\|^2 A$.
Using that fact, we get:
$\displaystyle \begin{array}{rcl}
{H \times E} & = & {\left( {H \times A} \right) - \alpha \left\| H \right\|^2 A} \\
{} & = & {\frac{1}
{\alpha }\left( {E - A} \right) - \alpha \left\| H \right\|^2 A} \\
{A + \alpha ^2 \left\| H \right\|^2 A} & = & {E - \alpha \left( {H \times E} \right)} \\ \end{array} $
The result follows at once.