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The equation of a plane sinusoidal wave propagating in the direction given by the unit vector $\mathbf{\hat k}$ is $$\mathbf E = \mathbf E_0 \cos\left(\frac{2\pi}{\lambda}\mathbf{\hat k} \cdot \mathbf r -\frac{2\pi}{T} t+\phi_0\right)$$ Here, $\mathbf{\hat k}$ is the unit vector at right angles to the wavefront, whose distance from the origin – as you'll see if you draw a diagram – is $\mathbf{\hat k} \cdot \mathbf r$. Writing, for convenience, $\mathbf k=\frac{2\pi}{\lambda} \mathbf{\hat k}$ and $\omega=\frac{2\pi}{T}$, $$\mathbf E = \mathbf E_0 \cos(\mathbf k \cdot \mathbf r -\omega t+\phi_0)$$ Expanding the dot product: $$\mathbf E = \mathbf E_0 \cos(k_xx+k_yy+k_zz -\omega t+\phi_0)$$ Comparing with the wave you are considering, we see that

$$k_x=a, k_y=0, k_z=b$$ So the components of $\mathbf{\hat k}$ are $$\frac {a\lambda}{2\pi}, 0, \frac {b\lambda}{2\pi}$$ The propagation direction therefore lies in the x-z plane, at an angle $\arctan{\tfrac ba}$ to the x axis.

We note that, since $\mathbf{\hat k}$ is a unit vector, $$\left(\frac {a\lambda}{2\pi}\right)^2+ 0^2+\left( \frac {b\lambda}{2\pi}\right)^2=1.$$

The equation of a plane sinusoidal wave propagating in the direction given by the unit vector $\mathbf{\hat k}$ is $$\mathbf E = \mathbf E_0 \cos\left(\frac{2\pi}{\lambda}\mathbf{\hat k} \cdot \mathbf r -\frac{2\pi}{T} t+\phi_0\right)$$ Here, $\mathbf{\hat k}$ is the unit vector at right angles to the wavefront, whose perpendicular distance from the origin – as you'll see if you draw a diagram – is $\mathbf{\hat k} \cdot \mathbf r$. Writing, for convenience, $\mathbf k=\frac{2\pi}{\lambda} \mathbf{\hat k}$ and $\omega=\frac{2\pi}{T}$, $$\mathbf E = \mathbf E_0 \cos(\mathbf k \cdot \mathbf r -\omega t+\phi_0)$$ Expanding the dot product: $$\mathbf E = \mathbf E_0 \cos(k_xx+k_yy+k_zz -\omega t+\phi_0)$$ Comparing with the wave you are considering, we see that

$$k_x=a, k_y=0, k_z=b$$ So the components of $\mathbf{\hat k}$ are $$\frac {a\lambda}{2\pi}, 0, \frac {b\lambda}{2\pi}$$ The propagation direction therefore lies in the x-z plane, at an angle $\arctan{\tfrac ba}$ to the x axis.

We note that, since $\mathbf{\hat k}$ is a unit vector, $$\left(\frac {a\lambda}{2\pi}\right)^2+ 0^2+\left( \frac {b\lambda}{2\pi}\right)^2=1.$$

The equation of a plane sinusoidal wave propagating in the direction given by the unit vector $\mathbf{\hat k}$ is $$\mathbf E = \mathbf E_0 \cos(\mathbf k \cdot \mathbf r -\omega t+\phi_0)\ \ \ \ \ \text {in which}\ \ \ \ \ \mathbf k=\frac{2\pi}{\lambda} \mathbf{\hat k}$$ Expanding the dot product: $$\mathbf E = \mathbf E_0 \cos(k_xx+k_yy+k_zz -\omega t+\phi_0)$$ Comparing with the wave you are considering, we see that

$$k_x=a, k_y=0, k_z=b$$ So the components of $\mathbf{\hat k}$, the unit vector in the propagation direction, are $$\frac {a\lambda}{2\pi}, 0, \frac {b\lambda}{2\pi}$$ The propagation direction therefore lies in the x-z plane, at an angle $\arctan{\tfrac ba}$ to the x axis.

We note that, since $\mathbf{\hat k}$ is a unit vector, $$\left(\frac {a\lambda}{2\pi}\right)^2+ 0^2+\left( \frac {b\lambda}{2\pi}\right)^2=1.$$

The equation of a plane sinusoidal wave propagating in the direction given by the unit vector $\mathbf{\hat k}$ is $$\mathbf E = \mathbf E_0 \cos\left(\frac{2\pi}{\lambda}\mathbf{\hat k} \cdot \mathbf r -\frac{2\pi}{T} t+\phi_0\right)$$ Here, $\mathbf{\hat k}$ is the unit vector at right angles to the wavefront, whose distance from the origin – as you'll see if you draw a diagram – is $\mathbf{\hat k} \cdot \mathbf r$. Writing, for convenience, $\mathbf k=\frac{2\pi}{\lambda} \mathbf{\hat k}$ and $\omega=\frac{2\pi}{T}$, $$\mathbf E = \mathbf E_0 \cos(\mathbf k \cdot \mathbf r -\omega t+\phi_0)$$ Expanding the dot product: $$\mathbf E = \mathbf E_0 \cos(k_xx+k_yy+k_zz -\omega t+\phi_0)$$ Comparing with the wave you are considering, we see that

$$k_x=a, k_y=0, k_z=b$$ So the components of $\mathbf{\hat k}$ are $$\frac {a\lambda}{2\pi}, 0, \frac {b\lambda}{2\pi}$$ The propagation direction therefore lies in the x-z plane, at an angle $\arctan{\tfrac ba}$ to the x axis.

We note that, since $\mathbf{\hat k}$ is a unit vector, $$\left(\frac {a\lambda}{2\pi}\right)^2+ 0^2+\left( \frac {b\lambda}{2\pi}\right)^2=1.$$

The equation of a plane wave propagating in the direction $\mathbf{\hat k}$ is $$\mathbf E = \mathbf E_0 \cos(\mathbf k \cdot \mathbf r -\omega t+\phi_0)\ \ \ \ \ \text {in which}\ \ \ \ \ \mathbf k=\frac{2\pi}{\lambda} \mathbf{\hat k}$$ Expanding the dot product: $$\mathbf E = \mathbf E_0 \cos(k_xx+k_yy+k_zz -\omega t+\phi_0)$$ Comparing with the wave you are considering, we see that

$$k_x=a, k_y=0, k_z=b$$ So the components of $\mathbf{\hat k}$, the unit vector in the propagation direction, are $$\frac {a\lambda}{2\pi}, 0, \frac {b\lambda}{2\pi}$$ The propagation direction therefore lies in the x-z plane, at an angle $\arctan{\tfrac ba}$ to the x axis.

We note that, since $\mathbf{\hat k}$ is a unit vector, $$\left(\frac {a\lambda}{2\pi}\right)^2+ 0^2+\left( \frac {b\lambda}{2\pi}\right)^2=1.$$

The equation of a plane sinusoidal wave propagating in the direction given by the unit vector $\mathbf{\hat k}$ is $$\mathbf E = \mathbf E_0 \cos(\mathbf k \cdot \mathbf r -\omega t+\phi_0)\ \ \ \ \ \text {in which}\ \ \ \ \ \mathbf k=\frac{2\pi}{\lambda} \mathbf{\hat k}$$ Expanding the dot product: $$\mathbf E = \mathbf E_0 \cos(k_xx+k_yy+k_zz -\omega t+\phi_0)$$ Comparing with the wave you are considering, we see that

$$k_x=a, k_y=0, k_z=b$$ So the components of $\mathbf{\hat k}$, the unit vector in the propagation direction, are $$\frac {a\lambda}{2\pi}, 0, \frac {b\lambda}{2\pi}$$ The propagation direction therefore lies in the x-z plane, at an angle $\arctan{\tfrac ba}$ to the x axis.

We note that, since $\mathbf{\hat k}$ is a unit vector, $$\left(\frac {a\lambda}{2\pi}\right)^2+ 0^2+\left( \frac {b\lambda}{2\pi}\right)^2=1.$$