A plane electromagnetic wave - phase change - amplitude 
A plane electromagnetic wave has the shape:
$\vec{E}(\vec{r},t)=E_0\cdot cos(\vec{k}\vec{r}-\omega t)\cdot \vec{e}_y$
$\vec{B}(\vec{r},t)=(B_1\cdot cos(\vec{k}\vec{r}-\omega t)+B_2\cdot sin(\vec{k}\vec{r}-\omega t))\cdot \vec{e}_z=B_0\cdot cos(\vec{k}\vec{r}-\omega t+\phi)\cdot \vec{e}_z$
In what direction is $\vec{k}$ facing?
Determine the amplitude $B_0$ and the phase change $\phi$ between $\vec{E}$ and $\vec{B}$.

We got this one in our lecture of experimental physics and it kind of bugs me because I can't find the right approach to this one.
How can I get from what is given to the direction of $\vec{k}$. I mean, I know that electromagnetic wave is moving along the x-axis from that, meaning $\vec{r}$ is too? Can I conclude the direction from the fact that there is a dot product between $\vec{k}$ and $\vec{r}$?
About the second part: In general $E$ and $B$ are in phase, right? Since that's not the case here, can I assume that it results from a reflection on a surface? But how would I get to $B_0$ and the phase change without explicit values for $\omega$ and such?
 A: 
How can I get from what is given to the direction of $\vec{k}$

The E-field has only a y component. And the B-field has only a z-component. Since this is a plane wave, then you know that the direction of propagation must be in either positive or negative x. 
From the information given, I believe it's actually ambiguous which direction the wave is travelling. 

In general E and B are in phase, right?

Yes. In a plane wave they are in phase. In a near-field radiation situation, they might not be. See further discussion in this electronics stackexchange question

Since that's not the case here, can I assume that it results from a reflection on a surface?

No, even if the plane wave resulted from a reflection off a surface, the E and B fields would be in phase.

But how would I get to $B_0$ and the phase change without explicit values for ω and such?

You can determine $B_1$ and $B_2$ from knowing that the E and B fields are in phase. Then you can determine $B_0$ and $\phi$ from those ($\phi$ = 0).
A: Enough time has passed that this homework-type question can be answered. It can be done in two ways. The first, by knowing the general properties of transverse electromagnetic waves; the second, by direct application of Maxwell's equations.
Method 1:
I will assume that the wave is travelling in a neutral, non-conducting medium, so that the wave-vector is a real number.
If we make this assumption then the E- and B-field should be in phase, so $\phi=0$.
The amplitude of the magnetic field is $E_0$ divided by the wave velocity. So $B_0 = E_0 k/\omega$.
The E-field, B-field and wave vector are mutually perpendicular and $\vec{E}\times \vec{B}$ should point in the direction of the wave vector. Since $\vec{e_y}\times \vec{e_z}=\vec{e_x}$, then $\vec{k} = k\vec{e_x}$ and $\vec{k}\cdot \vec{r} = kx$.
If $\phi=0$, then, since sine and cosine of the same argument are orthogonal functions, the wave only has a cosinusoidal component, so $B_1 = B_0$ and $B_2 = 0$.
Method 2:
Gauss's law in a neutral medium is that
$$\nabla \cdot \vec{E}=0\ .$$
If we let $\vec{k}=k_x \vec{e_x}+ k_y\vec{e_y}+ k_z\vec{e_z}$ and take the divergence of the electric field (which only has a y-component), then
$$ -k_y E_0\sin(\vec{k}\cdot \vec{r}-\omega t)=0$$
and so $k_y=0$.
We can then use Faraday's law to get an expression for the B-field.
$$\vec{B} = -\int \nabla \times \vec{E}\ dt\ .$$
$$\vec{B}= -\int -\vec{e_x} \frac{\partial E}{\partial z} + \vec{e_z}\frac{\partial E}{\partial x} \ dt$$
$$\vec{B} = \int E_0 \sin(\vec{k}\cdot\vec{r}-\omega t)(-k_z \vec{e_x}+ k_x\vec{e_z})\ dt, $$
$$\vec{B} = \frac{E_0}{\omega}\cos(\vec{k}\cdot\vec{r}-\omega t)(-k_z \vec{e_x}+ k_x\vec{e_z})\ .$$
Comparing this with
$$ B_0 \cos(\vec{k}\cdot\vec{r}-\omega t +\phi)\vec{e_z}$$
we can immediately see that $k_z=0$, $\phi=0$ and that $B_0 = E_0 k_x/\omega$.
Note, that if $k$ were a complex number (for example, a wave in a conductor), then $B_0$ would also be complex. To have $B_0$ as a real number, the phase angle $\phi$ would have to be a non-zero value.
A: Recall that $cos\alpha\cdot sin\beta +cos\beta\cdot sin\alpha = cos(\alpha+\beta)$. 
If you suppose that $B_1=B_0\cdot sin\phi$ and $B_2=B_0\cdot cos\phi$, it turns out that:
$B_1cos(kr-\omega t)+B_2cos\phi=B_0\cdot sin\phi \cdot cos(kr-\omega t)+B_0\cdot sin(kr-\omega t)\cdot cos\phi=$
$=B_0\cdot cos(kr-\omega t+\phi)$.
