# Boundary condition for E field

My book says that the boundary condition for the E field is: $$\hat{n} \times (\textbf{E}_1 - \textbf{E}_2) = 0$$

and then concludes that the above condition can be summarized by the statement, "The tangential electric field is continuous across the boundary surface."

This is probably something very simple I'm missing, but why is that statement equivalent to the above expression?

The tangential electric field is just the electric field projected onto the surface. So the requirement of continuity means that the projections of $\mathbf{E_{1}}$ and $\mathbf{E_{2}}$ onto the surface must be equal, which is what the equation describes.
If the surface in question is the $xy$-plane, and $\mathbf{\hat n}$ is oriented in the $z$-direction, then the cross product is nonzero if $\mathbf{E}_{1} - \mathbf{E}_{2}$ has either $x$ or $y$ components. The condition that the cross product be zero requires that $E_{1, x} = E_{2, x}$ and $E_{1, y} = E_{2, y}$. If only their $z$ component differs, then $\mathbf{E}_{1} - \mathbf{E}_{2} = (E_{1, z} - E_{2, z})\mathbf{\hat z}$. The cross product of two vectors with the same orientation is zero, so $\mathbf{\hat n} \times (\mathbf{E}_{1} - \mathbf{E}_{2}) = 0$ and the boundary condition is satisfied.