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.

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.

$$ 2A\cos(k x)\cos(\omega t)= A[\cos(kx -\omega t)+\cos(-kx - \omega t)] $$

In a scalar formulation, it's convenient to define the positive and negative direction of propogation through the negative $\omega t$. Since we'll be working in a vector formulation with $k$, it's easier to show the direction through the sign on $k$. 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 equal magnitude, but opposite direction. This suggests that two plane waves with common phase, that have the same amplitude of wavenumber, will in fact make a standing wave when viewed within a particular plane.

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 wavenumbers must be equal to get any kind of standing wave, not just individual components.

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.