# Question in reflection and transmission at oblique incidence.

I am reading Griffiths p387. It is my understanding that
$$\tilde E(\vec r,t)=\hat r E_0 e^{j(\omega t-kr)}$$
Where ##\vec r =\hat x x+\hat y y+ \hat z z## is the positional vector from the origin to the observation point ( x,y,z).
$$\Rightarrow\;\tilde E(\vec r,t)=\hat r E_0 e^{-jkr}\;=\;\left(\frac {\hat x x+\hat y y+ \hat z z}{\sqrt{x^2+y^2+z^2}}\right) E_0e^{j[-(\hat x k_x+\hat y k_y+\hat z k_z)\cdot(\hat x x+\hat y y+\hat z z)]}\;=\; \left(\frac {\hat x x+\hat y y+ \hat z z}{\sqrt{x^2+y^2+z^2}}\right) E_0e^{-j[( x k_x+ y k_y+z k_z)]}$$

In Griffiths, he let the incident TEM wave travels in ##\vec k_I## direction. So he let
$$\vec E_I(\vec r,t)= \vec E_{0I} e^{j(\omega t - \vec k_I\cdot \vec r)},\;\vec E_R(\vec r,t)= \vec E_{0R} e^{j(\omega t - \vec k_R\cdot \vec r)},\;\vec E_T(\vec r,t)= \vec E_{0T} e^{j(\omega t - \vec k_T\cdot \vec r)},\;$$
To expand one out:
$$\vec E_I(\vec r,t)= \vec E_{0I} e^{j(\omega t - \vec k_I\cdot \vec r)}=\hat x E_{0I}e^{j[\omega t - (xk_{Ix}+yk_{Iy}+zk_{Iz})]}$$

I have a problem with this, as you can see from the scanned page, the direction of the incident, reflected and transmitted wave is in direction of ##\vec k_I,\;\vec k_R\;\hbox { and } \;\vec k_T##. But he gave all three as ##\vec E(\vec r,t)##. This mean all three are in ##\vec r## direction. That is not right.
Later, he actually equated
$$xk_{Ix}=xk_{Rx}\;\Rightarrow\;k_{Ix}=k_{Rx}$$
This mean he used the same ##\vec r## in all three, that is questionable. I am not saying the final result is wrong, just the representation is questionable.

Thanks

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