I think partial of this question being asked before but I have some other doubts.As shown in this post how to add two plane waves if they are propagating in different direction?, reading the third reply by Wouter. If we have two waves
$$ f_1 = \sin(\vec{k}\cdot\vec{r} + \omega t), f_2 = \sin(\vec{q}\cdot\vec{r} - \omega t) $$
1) my first question, in one text, I saw that to create the standing wave, we should have the two sinusoidal waves propagate in counter direction. I am confusing with above function of wave. I know that the group velocity is defined as $$\vec{v}_p = \hat{k}\omega/|\vec{k}|$$ so if we want two waves propagate in opposite direction, we could have one $\omega$ and the other $-\omega$ or we have have one $\vec{k}$ positive and the other one $-\vec{k}$. But starting from above waves, $\omega$ is opposite, so we should make sure $\vec{k}=\vec{q}$. But what's the physical significance to make $-\omega$? To me, it makes more sense to
$$ f_1 = \sin(\vec{k}\cdot\vec{r} + \omega t), f_2 = \sin(\vec{q}\cdot\vec{r} + \omega t) \quad \mbox{with} \quad \vec{q}=-\vec{k} $$
2) Let start from the last formulas, let $\vec{q}=-\vec{k}$ so to have the sum in the form
$$ f_1+f_2 = \sin(\vec{k}\cdot\vec{r})\cos(\omega t) $$
what I understand from the text is standing wave should be stationary in space. But there is a modulated amplitude $\cos(\omega t)$ there so does it really stationary (or standing wave)?
3) If I have two such "standing waves" propagating along different direction, says $\cos^{-1}(\vec{p}\cdot\vec{q})=\pi/6$
$$ s_1 = \sin(\vec{p}\cdot\vec{r})\cos(\omega t), \quad s_2 = \sin(\vec{q}\cdot\vec{r})\cos(\omega t) $$
so will these two wave makes a two-dimensional standing wave if $|q|=|p|$?