The question states that the direction of propagation of E is +x and +z plane. The goal here is to find the reflected component of E and H. I know that the propagation changes sign once it hits the boundary. but what i don't understand in part a) why z component changes sign not x? why not both? Also in part b) I cannot visualize how the the sin and cosine component works here, also why these are added to reflected component of H ? can anyone draw this out for me please and explain it? Thanks
I took your picture from your other question and I edited to make it match to your problem.
As you can see the wave progresses in the [3,0,4] direction. This is the argument of the complex exponential. Remember than ax+by+cz=K (K is a constant) would be a plane and the vector perpendicular to it is [a,b,c].
Vector \$\vec \beta _i\$ is therefore [3,0,4]. You can see from the picture that the reflected \$\vec \beta _r\$ is [3,0,-4]. It reflects on the z=0 plane (and that's why z changes sign).
In the other hand, the cos and sin components allow to write the vector from its polar coordinates.
If you look at \$\vec H _i\$ it is easy to verify that its angle with \$\beta _z\$ is \$\theta_i\$ therefore its coordinates are \$[-H_i \cos(\theta_i) , 0, H_i sin(\theta_i)]\$.
Because \$\Gamma\$ is negative we have a direction change for the reflected E field. Also E, H and propagation direction must follow the 3 fingers rule, before and after the reflection, as they do in the figure.
For \$\vec H _r\$ you can see from the figure that its coordinates will be \$[-H_r \cos(\theta_r) , 0, -H_r sin(\theta_r)]\$ as in the solution.