Laplacian

Needs["VectorAnalysis`"]

Now let's use a description of displacement for a plane wave, in any direction, 
where n is the vector that describes the direction in which the wave is travelling, 
k is the wavenumber scalar, (Subscript[u, p]) describes the direction of particle motion 
during propagation and x is the location of a point in space through 
which the wave traverses.  Only the original amplitude at the source ((Subscript[A, 0])) , 
the velocity of propagation and the time are scalar values. The rest are vectors in 3D.
|k|= (1/(2ℼλ))with units of (m^(-1))

(OverVector[n]) = {(Subscript[n, 1]), (Subscript[n, 3]), (Subscript[n, 3])}
(Subscript[(OverVector[u]), p]) = {(Subscript[u, p1]), (Subscript[u, p2]), (Subscript[u, p3])}
(OverVector[x]) = {(Subscript[x, 1]), (Subscript[x, 2]), (Subscript[x, 3])}
Clear[V,t,(Subscript[A, 0])]

(OverVector[u])   = (Subscript[A, 0])(Subscript[(OverVector[u]), p]) Exp[I k ((OverVector[n]).(OverVector[x]) - V t)]

Let' s try to visualize this function in two dimensions, by making the propagation vector (n):
the x axis, the direction of particle motion ((Subscript[u, p])) : the y axis. Now consider what is happening
at the location where x = ( 1,0,0) during the passage of time.  Make the wavelength be L in
only the x direction.

(Subscript[A, 0]) = 1;
f = 1      ;              (* frequency in Hz*)
V =  1000     ;          (* velocity in m/s*)
ω   = 2Pi f    ;     (* frequency in radians per second*)
L   = V/  ω;

(OverVector[n]) = {0, 1, 0};
k= 1/L;
(Subscript[(OverVector[u]), p]) = {0, 1, 0}
(OverVector[x]) = {X, 0, 0}

(OverVector[u])   = (Subscript[A, 0]) (Subscript[(OverVector[u]), p]) Exp[I  k((OverVector[n]).(OverVector[x]) - V t)];
ExpToTrig[(OverVector[u])];
Plot3D[ Re[(OverVector[u]) ], {t,0,5}, {X,-100,100}, PlotRange->{{0,5},{-100,100}},AxesLabel->Automatic]