Why does $\mathbf{v} \cdot \nabla H =0$ imply that $H$ is constant along a stream line? Context
I am trying to follow a derivation of Bernoulli's equation from Acheson's Elementary Fluid Dynamics 1990.
After making some simplifying assumptions and doing some algebra I got to:
$$( \nabla \times \mathbf{v}) \times \mathbf{v} = - \nabla H$$
where:
$\mathbf{v}$ is a vector field representing the fluid velocity
$H$ is an expression (we are trying to show that $H$ is a constant).
Next we take the dot product with $\mathbf{v}$.
$$\mathbf{v}\cdot (( \nabla \times \mathbf{v}) \times \mathbf{v}) =  -\mathbf{v} \cdot \nabla H$$
The left hand side is zero (this can be shown using vector identities).
Hence we get to $-\mathbf{v} \cdot \nabla H =0$ which implies that $\mathbf{v} \cdot \nabla H =0$
Apparently it is now trivial to see that "$H$ is constant along the streamlines". But I do not understand why?
Question

Why does $\color{blue}{\mathbf{v} \cdot \nabla H =0}$ imply that $\color{blue}{H}$ is constant along a stream line?


Here is an example fluid velocity field with streamlines if anyone wants to refer to a concrete example:
$\mathbf{v} = x\hat{i}+(-y)\hat{j}+ 0\hat{k}$

here is the same field with the corresponding potential shown as contours (which I notice are perpendicular to the stream lines (the dot product of perpendicular vectors is zero...)).

 A: $\mathbf{v} \cdot \nabla H$ is the directional derivative of $H$ in the direction of $\mathbf{v}$. Since $\mathbf{v} \cdot \nabla H =0$, we have that the instantaneous rate of change of $H$ in the direction of $\mathbf{v} $ is $0$. So $H$ is constant in the direction of $\mathbf{v}$ and thus $H$ is constant along the streamlines of the fluid.
A: The chain rule also may shed some light. 
$\vec{v}$ is the velocity vector so:
$\vec{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k} =\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k}$
$$\nabla H=\frac{\partial H}{\partial x}\hat{i}+\frac{\partial H}{\partial y}\hat{j}+\frac{\partial H}{\partial z}\hat{k}$$
By the chain rule we have:
$$\frac{dH}{dt}=\frac{\partial H}{\partial x}\frac{dx}{dt}+\frac{\partial H}{\partial y}\frac{dy}{dt}+\frac{\partial H}{\partial z}\frac{dz}{dt}=\frac{d\vec{s}\cdot \nabla H}{dt}$$
And keep in mind, total derivative of $f$ along a path is $df=\nabla f\cdot d\vec{s}$
where $d\vec{s}$ is infinitesimal length element in direction of change. 
