1. ## help with vectors

Hello every one I have this vector relation
E=A+$\alpha$(HXA) where E,A and H are vectors I want to proof that
A=(E-$\alpha$(HXA))/$\sqrt{1+H^{2}\alpha^{2}}$
any one remeber how we can do this using vector identities or simple geometry or both of them (may be)?

2. Originally Posted by omeganeu
Hello every one I have this vector relation
E=A+$\alpha$(HXA) where E,A and H are vectors I want to proof that
A=(E-$\alpha$(HXA))/$\sqrt{1+H^{2}\alpha^{2}}$
any one remeber how we can do this using vector identities or simple geometry or both of them (may be)?
As stated this is not true.
It is true that $A=E- \alpha(H \times A)$.
Because H is a vector $H^2$ has no meaning.

3. H is the vector and H is its magnitude

actually I found this relation in one book but I do not remember how I can get it

4. the correct equation is A=(E-$\alpha$(HXE))/(1+$\alpha$$^{2}$H$^{2}$) I am really sorry for that mistake

5. Originally Posted by omeganeu
the correct equation is A=(E-$\alpha$(HXE))/(1+$\alpha$$^{2}$H$^{2}$) I am really sorry for that mistake
Even with the correction it still not true unless we know that $A~\&~H$ are perpendicular.
Did you leave that out?

6. actually E and H should be perpendicular to each other. E has two component in the x-and the y- axis and H is parallel to the z-axis

7. Originally Posted by omeganeu
actually E and H should be perpendicular to each other.
If that is the case then clearly $A~\& ~H$ are also perpendicular. Form which we get using a well known fact
$H \times \left( {H \times A} \right) = \left( {H \cdot A} \right)H - \left( {H \cdot H} \right)A = - \left\| H \right\|^2 A$.
Using that fact, we get:
$\begin{array}{rcl}
{H \times E} & = & {\left( {H \times A} \right) - \alpha \left\| H \right\|^2 A} \\
{} & = & {\frac{1}
{\alpha }\left( {E - A} \right) - \alpha \left\| H \right\|^2 A} \\
{A + \alpha ^2 \left\| H \right\|^2 A} & = & {E - \alpha \left( {H \times E} \right)} \\ \end{array}$

The result follows at once.

8. yes finally I get it! thank you very much for the great help and sorry for the mistakes.