how to add two plane waves if they are propagating in different direction?

In the undergraduate course about the wave, there stated for two harmonic waves propagating in opposite direction, then the resulting wave will be a standing wave. In math, it is like

$$y_1 = A\sin(kx + \omega t), y_2 = A\sin(kx - \omega t)$$

so

$$y = y_1+y_2 = A\sin(kx + \omega t) + A\sin(kx - \omega t) = 2A \sin(kx)\cos(\omega t)$$

I am thinking what happen if we have the two plane waves propagating along two different direction (says making angle 60 degree, i.e. the two wave are making ). I know that if that's the case, we cannot write $kx$ but we need to consider the $k$ is a vector such that

$$y_1 = A\sin(\vec{k}\cdot\vec{r} + \omega t), y_2 = A\sin(\vec{k}\cdot\vec{r} - \omega t)$$

But if we look at the horizontal direction (i.e. x) and vertical direction (i.e. y), what can we tell about the resulting wave along x and along y? I am thinking from physical point of view, if we look at the horizontal direction, should the waves still added up to a standing wave because the x components of waves are propagating in opposite direction. But along the vertical direction, the y components of waves are propagating in the same direction so there is no standing wave. Is that correct? If so, how to prove that in math? The term $\vec{k}\cdot\vec{r}$ is very confusing!

Answer

The question is unclear, but I believe can be summarized as "can standing waves form from plane waves that propagate at some arbitrary angle to each other?"

A standing wave is most easily understood in one dimension, and can be described by the equation.

$$u = A\cos(k x)\cos(\omega t)$$

It's a simple product-sum trig identity, which can be found on this page that relates the standing wave to the waves propagating in opposite directions.

$$2A\cos(k x)\cos(\omega t)= A[\cos(kx -\omega t)+\cos(-kx - \omega t)]$$

In a scalar formulation, it's convenient to define the positive and negative direction of propogation through the negative $\omega t$. Since we'll be working in a vector formulation with $k$, it's easier to show the direction through the sign on $k$. There could also be an arbitrary phase.

Now to show that something like a standing wave could occur in two dimensions (easily generalized to 3 dimensions), it's easier to use complex exponentials to represent the waves. Adding the two waves (q is the wavenumber of the second wave):

$$Ae^{i(\mathbf{k \cdot r}-\omega t)} + Ae^{i(\mathbf{q \cdot r} - \omega t)} = Ae^{ik_{y}y}e^{i(k_{x}x-\omega t)}+Ae^{iq_{y}y}e^{i(q_{x}x-\omega t)}$$ If the y component of the wavenumber is identical for both waves, then the y component can be combined with the amplitude to form a complex amplitude common between both waves, with a phase that depends on y. $$Ae^{ik_{y}y}(e^{i(k_{x}x-\omega t)}+e^{i(q_{x}x-\omega t)})$$ Reverting back to a trig representation, and ignoring the y dependent phase: $$A[\cos(k_{x}x-\omega t)+cos(q_{x}x-\omega t)]$$ You should be able to recognize that this is a standing wave if the x components of the wavenumber are equal magnitude, but opposite direction. This suggests that two plane waves with common phase, that have the same amplitude of wavenumber, will in fact make a standing wave when viewed within a particular plane.