The question is unclear, but I believe can be summarized as "can standing waves form from plane waves that propagate at some arbitrary angle to each other?"
A standing wave is most easily understood in one dimension, and can be described by the equation. $$ u = A\cos(k x)\cos(\omega t) $$
It's a simple product-sum trig identity, which can be found on this page that relates the standing wave to the waves propagating in opposite directions.
$$ 2A\cos(k x)\cos(\omega t)= A[\cos(kx-\omega t)+cos(kx+\omega t)] $$
There could also be an arbitrary phase.
Now to show that something like a standing wave could occur in two dimensions (easily generalized to 3 dimensions), it's easier to use complex exponentials to represent the waves. Adding the two waves (q is the wavenumber of the second wave): $$ Ae^{i(\mathbf{k \cdot r}-\omega t)} + Ae^{i(\mathbf{q \cdot r} + \omega t)} = Ae^{ik_{y}y}e^{i(k_{x}x-\omega t)}+Ae^{iq_{y}y}e^{i(q_{x}x+\omega t)} $$ If the y component of the wavenumber is identical for both waves, then the y component can be combined with the amplitude to form a complex amplitude common between both waves, with a phase that depends on y. $$ Ae^{ik_{y}y}(e^{i(k_{x}x-\omega t)}+e^{i(q_{x}x+\omega t)}) $$ Reverting back to a trig representation, and ignoring the y dependent phase: $$ A[\cos(k_{x}x-\omega t)+cos(q_{x}x+\omega t)] $$ You should be able to recognize that this is a standing wave if the x components of the wavenumber are also equal. Ultimately, the waves must be propagating in opposite direction with the same wavenumber to form a standing wave just like in the one-dimensional case.