Perpendicular complex-valued vectors

Question

I'm reading an E&M Textbook and somewhat confused about how to find perpendicular vectors that satisfy the right hand rule, when the coefficients are complex. For example, if $E$ has direction $(\hat{x}+j\hat{z})$ and the wave propogates in the $+\hat{y}$ direction, then the book says $H$ has direction $(j\hat{x}-\hat{z})$. This should satisfy the right hand rule, so that $E\times H$ is in the $+\hat{y}$ direction. But I'm not sure how to derive the direction of $H$ from $E$. When the vectors are real-valued I can draw a diagram, but when they're complex I have no intuition.

Answer 1

There's a few ways to look at this.

First, since this arising from a physics problem, we can think physically about what these directions really mean. $\hat{x} + j\hat{z}$ means that the electric field is initially pointing in the $\hat{x}$ direction and then quarter of a period later (90 degrees phase) it points in the $\hat{z}$ direction. When the electric field points in the $\hat{x}$ direction and the wave is travelling in the $\hat{y}$ direction, then the magnetic field points in the $-\hat{z}$ direction. Later, when the electric field points in the $\hat{z}$ direction, the magnetic field must point in the $\hat{x}$ direction. In short, when the phase is zero, the magnetic field direction is $-\hat{z}$ and at 90 degree phase, magnetic field points in $\hat{x}$ direction, so the $H$ must point in $-\hat{z} + j\hat{x}$.

We can also intuitively come up with answer by realizing that if there are three mutually orthogonal vectors $A$, $B$, and $C$, then we can always find the direction of the third vector by taking cross products of the other two because the cross product of two vectors yields a vector that is orthogonal to both vectors. In particular, if $E \times H$ points in $\hat{n}$, then $\hat{n} \times E$ points in the direction of $H$.

We can algebraically show that this intuition is correct. Recall that

$$A \times (B \times C) = B(A\cdot C) - C (A\cdot B)$$ Letting $A = B = E$ and $C = H$, we see that

$$\hat{n} \times E \propto (E \times H) \times E \propto H$$ since $E$ is orthogonal to $H$ (proportional since the vectors aren't necesarily normalized). In other words, we can always get the direction of the $H$-field by computing $\hat{n} \times E$.