calculus - Intuition behind the directional derivative being zero when gradient is perpendicular to the direction

Suppose we have a multivariable function $f$ and I want to find it's directional derivative along the direction $\vec u$. Then the formula tells us that the directional derivative will be $$\nabla f\cdot \vec u = |\nabla f||u|cosθ$$ where $θ$ is the angle between these two vectors. That means that, for the directional derivative to be $0$, then $cosθ = 0 \to θ= π/2$.

Is there an intuitive way to understand this? Why the gradient needs to be perpendicular to my chosen direction in order for the directional derivative, i.e.the slope of the function in the particual point along my chosen direction, to be $0$ ?

Thanks in advance.

EDIT : I can maybe understand something related to the contours. https://en.wikipedia.org/wiki/Directional_derivative The first picture here shows a contour plot, where we see that the gradient is perpendicular to the contours and we also see a dir. derivative. If these 2 vectors were perpendicular,then the dir. derivative would have to be tangent to the contour and therefore, our unit vector $u$ would be tangent to it. That means that our direction is tangent to the contour. So for small steps, the function wouldn't change value. So our rate of change would be zero, i.e. the dir. derivative would be zero. Is that correct?