How to find the characteristics of a wave (which is traveling along a straight line in the $x$-$y$ plane) from the given equation? If have dealt with equations of the wave which are travelling along the coordinate axis. But now I found a question which asks about a wave which travels in the x-y plane, but not along the coordinate axis:

A plane progressive harmonic wave is given by the equation:
\begin{align}\phi=\phi_m sin(\omega t-3x+4y+\frac\pi 3) \end{align}
where x and y are in meters, and t is in seconds. If v is the speed of wave w.r.t. the wave medium, then
{i} Write the unit vector in the direction of propagation of the wave
{ii} Write the value of wave constant k
{iii} Find the speed v of the wave.

This question is to give an idea of what I want to ask. I don't know how to differentiate between $\vec{k}$ and $\vec{r}$ in the general wave equation:$$\phi=sin(\omega t-\vec{k}\cdot\vec{r})$$
Can someone explain this.
 A: Wavenumber $k$ is a point.

*

*$\vec k = (3,-4)$


*$k = |\vec k| = \sqrt{(3)^2 + (-4)^2} =5$


*$v = \frac{w}{k} = \frac{w}{5}$
A: Plane waves have the general form
$$\sin(\omega t-\vec k\cdot \vec r+\phi_0)$$
where $\omega$ is the angular frequency, $\vec k$ is the wave vector, $\vec r=(x,y)^T$ and $\phi_0$ is some phase offset. From this you should be able to easily read off $\vec k$.
(i) How do we interpret the direction of $\vec k$? The equation $\vec k\cdot \vec r=0$ describes a wavefront (or more generally: $\vec k\cdot\vec r=c$). Every point $\vec r$ that satisfies this equation has the same phase. The equation $\vec k\cdot \vec r=0$ is just a line perpendicular to $\vec k$ so all wavefronts are perpendicular to $\vec k$. The direction of propagation is perpendicular to all wavefronts so the direction of propagation is parallel to $\vec k$.
(ii) $k$ is just the length of $\vec k$.
(iii) Imagine a point $\vec r$ on a wavefront which has $\omega t-\vec k\cdot \vec r=c$ at time $t$. After a small timestep $\Delta t$ this wavefront has moved along $\vec k$ by a distance which I will call $\Delta r$. Since it is the same wavefront we should still have
$$\omega(t+\Delta t)-\vec k\cdot(\vec r+\Delta r\hat k)=c.$$
We can now solve for the speed $\Delta r/\Delta t$:
\begin{align}
\omega(t+\Delta t)-\vec k\cdot(\vec r+\Delta r\hat k)&=c\\
\underbrace{\omega t-\vec k\cdot\vec r}_{c}+\omega\Delta t-k\Delta r&=c\\
\omega\Delta t-k\Delta r&=0\\
\implies v=\frac{\Delta r}{\Delta t}&=\frac{\omega}{k}
\end{align}
Bonus: you can also define the vector $\vec\lambda=\frac{2\pi}{|\vec k|^2}\vec k$. The length of this vector is the wavelength, so this vector has the nice geometrical interpretation that it points from wavefront to wavefront.
