Consider an electron fired at a target. Taking the axis of motion to be $x$, and position to be $(x,y,z)$ then
$\Delta y = \Delta z = 0$
Therefore by the uncertainty principle
$\Delta p_y = \Delta p_z = \infty$
So the electron will spread out in $y$ and $z$ and never hit its target (unless that target was very close i.e. on the same scale as the initial effective size of the electron). How can it hit a target in reality?
I realise there is a possible duplicate to this question: How can particles travel in a straight line? The accepted answer for the duplicate is "because $\hbar$ is really small so the particles don't spread out much before hitting the target". I'm not sure this stands though. The time-dependent Schrodinger equation is:
$i\hbar\frac{\delta}{\delta t}\psi(r,t) = [-\frac{\hbar^2}{2\mu}\nabla^2 + V(r,t)]\psi(r,t)$
In the absence of external potential $V$ and dividing both sides by $\hbar$:
$i\frac{\delta}{\delta t}\psi(r,t) = -\frac{\hbar}{2\mu}\nabla^2\psi(r,t)$
So if we change $\hbar$ (or $\mu$), sure the particle will spread out less as a function of time, but won't it also move slower in exactly the same proportion to its lack of spreading out, so it still won't hit the target?