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Given at an interface $z=0$ : $$E = (30 \hat{x} + 20 \hat{y} + 10 \hat{z})\cos(\omega t)$$ And the direction of propagation is $z$. I want to find the normal electric field $(E_N)$ and the tangential electric field $(E_t)$

I know it is obvious that $E_t = (30\hat{x} + 20\hat{y})cos(\omega t)$ and $E_N = (10\hat{z}) cos(\omega t)$

But I am confused why doesn't $E \times \hat{n}$ work?

$$(30 \hat{x} + 20 \hat{y} + 10 \hat{z}) \cos(\omega t) \times \hat{z} $$

$$= (30\hat{y} -20\hat{x}) \cos\omega t$$?

The boundary conditions say (for example) at the interface of a conductor, the tangential component of the electric field is zero and writes $$\hat{n} \times \vec{E} = 0$$ But $\hat{n} \times \vec{E}$ does not get the tangential component of E (In the above example) Why?

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    $\begingroup$ I would say that directly taking the normal component and taking a cross product is not equivalent. Nevertheless, if there is no normal component, the cross product should yield 0. The reason is simply, if there is a non-zero magnitude normal component, there could be infinitely many vectors that could yield $\vec{0}$ when dot product with $n$. However, if there is none, that is, the magnitude is $0$, the only possible vector is $0$, which is unique. $\endgroup$
    – by24
    Commented Nov 14 at 7:16
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    $\begingroup$ $A \times B = 0$ iff A and B are colinear, meaning here $E$ is colinear to $\hat{n}$ - so it is colinear to the normal of the surface, and thus doesn't have a tangential component $\endgroup$
    – Ofek Gillon
    Commented Nov 14 at 7:58
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    $\begingroup$ However, remember that $E\times \hat{n}$ is perpendicular to $E$ as well, and thus the vector is not the tan. component of $E$ $\endgroup$
    – Ofek Gillon
    Commented Nov 14 at 7:59
  • $\begingroup$ @OfekGillon Yes that is exactly what I mean, it is not the tangential component of E then why references on the subject say sentences like: "The tangential components of E are equal --> E cross n = 0" $\endgroup$
    – Nero
    Commented Nov 14 at 8:58
  • $\begingroup$ Wait I think I understood it, if someone can confirm: It is actually that we know Et1 = Et2 (I.e. they tangential components cancel each other) and we want to express it mathematically using $E \times n = 0 $ (Which as you said means no tangential component, but won't simplify down to Et1 = Et2 right?) (I mean it implicity implies Et1-Et2 = 0 but expanding it won't get this exact equation. $\endgroup$
    – Nero
    Commented Nov 14 at 9:01

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A cross product should be sus, because it produces an axial vector, and setting a regular vector equal to an axial vector is "bad", unless your name rhymes with "Wu"$^1$.

So, given a normal $\hat{n}$, you have its projection operator:

$${\rm proj}_n = \hat n\hat n $$

so applying that to $\vec E$, you get, in index notation:

$$ \vec E_{\parallel} \rightarrow (n_i n_j) E_j$$

The left-over (complement) is the vector rejection:

$$ \vec E_{\perp} = \vec E - \vec E_{\parallel} \rightarrow (\delta_{ij} - n_i n_j) E_j $$

which is very different from an operator made from the cross product, such as:

$$ E_i \rightarrow [\epsilon_{ijk}n_k] E_j $$

For completeness, you can also get the vector reflection of $\hat n$, for which there is no cute subscript, nor unambiguous single operator symbol afaik:

$${\rm Ref}^{(n)}_{ij} = \delta_{ij} - 2n_in_j $$

[1] Beware parity violating [https://en.wikipedia.org/wiki/Wu_experiment] expressions when using the cross product

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