For a plane electromagnetic wave propagating in a dielectric medium, the electric field vector ($ \vec{E}$), the magnetic field vector ($\vec{B}$), and the wave vector ($\vec{k}$) are mutually perpendicular. The direction of wave propagation is given by the direction of $\vec{E} \times \vec{B}$, which is the same as the direction of $ \vec{k}$. The given electric field is $\vec{E} = (2\hat{i} - 1.5\hat{j})e^{i[k_0(3x+4y)-\omega t]} $. The term $k_0(3x+4y)$ represents $\vec{k} \cdot \vec{r}$. This indicates that the wave vector $\vec{k}$ is in the direction $3\hat{i} + 4\hat{j}$. The direction of $\vec{E}$ is given by the amplitude vector $(2\hat{i} - 1.5\hat{j})$. Let this be $\vec{E}_0$. We need to find a unit vector $\hat{B}$ such that $\vec{E}_0 \times \hat{B}$ is in the direction of $3\hat{i} + 4\hat{j} $. Let's check option A: $\hat{B} = -\hat{k}$. $ \vec{E}_0 \times \hat{B} = (2\hat{i} - 1.5\hat{j}) \times (-\hat{k}) = (2\hat{i} \times -\hat{k}) + = 1.5\hat{i} + 2\hat{j}$. The direction of this vector $1.5\hat{i} + 2\hat{j}$ is indeed the same as $3\hat{i} + 4\hat{j}$ (since $1.5\hat{i} + 2\hat{j} = \frac{1}{2}(3\hat{i} + 4\hat{j}) \times 2 = 3\hat{i} + 4\hat{j}$). Thus, with $\hat{B} = - \hat{k}$, $\vec{E}_0 \times \hat{B}$ is in the correct direction of wave propagation. Also, check perpendicularity: $\vec{E}_0 \cdot \hat{B} = (2\hat{i} - 1.5\hat{j}) \cdot (-\hat{k}) = 0$ and $(3\hat{i} + 4\hat{j}) \cdot (-\hat{k}) = 0$, confirming $\vec{E}$ and $\vec{k}$ are perpendicular to $\vec{B}$. Options B, C, and D are not unit vectors, and therefore cannot be the unit vector $\hat{B}$.