# I'm given that vec(E)=(mu_0 p_0 omega^2)/(4 pi r) [cos u(hat(x)-x/r hat(r))+sin u(hat(y)-y/r hat(r))] implies -(mu_0 p_0 omega^2)/(4 pi r) [cos u hat(x) xx hat(r) + sin u hat(y) xx hat(r)]=(1)/(c) hat(r) xx vec(E)

I'm given that
$\stackrel{\to }{E}=\frac{{\mu }_{0}{p}_{0}{\omega }^{2}}{4\pi r}\left[\mathrm{cos}u\left(\stackrel{^}{x}-\frac{x}{r}\stackrel{^}{r}\right)+\mathrm{sin}u\left(\stackrel{^}{y}-\frac{y}{r}\stackrel{^}{r}\right)\right]$
implies
$-\frac{{\mu }_{0}{p}_{0}{\omega }^{2}}{4\pi r}\left[\mathrm{cos}u\stackrel{^}{x}×\stackrel{^}{r}+\mathrm{sin}u\stackrel{^}{y}×\stackrel{^}{r}\right]=\frac{1}{c}\stackrel{^}{r}×\stackrel{\to }{E},$
but I don't follow how to get from the former to the latter.
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berzamauw
You get the second equation by taking the cross product of the first one with $\frac{1}{c}\stackrel{^}{r}$ on the left and remembering that $\stackrel{\to }{x}×\stackrel{\to }{x}=\stackrel{\to }{0}$ , so in particular $\stackrel{^}{r}×\stackrel{^}{r}=\stackrel{\to }{0}$ ; and also that $\stackrel{^}{r}×\stackrel{^}{x}=-\stackrel{^}{x}×\stackrel{^}{r}$