Why is stress a scalar quantity, even though mathematically, it is (Internal restoring force/Area)? Basically, what I need to know guys is that when we divide a vector by a scalar, we get a vector. Then what is different in the case of stress?? I mean, WHY IS IT STILL A SCALAR? 
 A: Stress is a rank-two tensor that couples one 3-D vector (a direction associated with an area) to another (a force). The tensor can hold 3×3=9 values according to the various directional combinations, but for static problems, only six values can be independent.
In symbolic form, the stress $\boldsymbol\sigma$ is
$$\boldsymbol\sigma=\begin{bmatrix}\sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ 
\sigma_{yx} & \sigma_{yy} & \sigma_{yz}\\\sigma_{zx} & \sigma_{zy} & \sigma_{zz}\end{bmatrix}=\begin{bmatrix}\sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ 
& \sigma_{yy} & \sigma_{yz}\\& & \sigma_{zz}\end{bmatrix}$$
where the missing elements are the nonindependent ones for a nonaccelerating infinitesimal cube. (A free-body diagram would show you that if $\sigma_{xy}\neq\sigma_{yx}$, then the cube must start to rotate.) One of the indices (it doesn't matter which due to the symmetry) is the direction of the surface normal vector, and the other is the force direction. Again, in the static case, $\sigma_{xx}=\sigma_{(-x)(-x)}$.
Pressure, a scalar, is one-third the negative of the trace of the stress tensor written in matrix form: $P=-\frac{1}{3}(\sigma_{xx}+\sigma_{yy}+\sigma_{zz})$. This is one of three invariants (i.e., coordinate-system-independent parameters) of the stress tensor.
If you were considering a single scalar value of stress, it's because you'd already implicitly chosen the surface orientation and force direction.
A: Good question. Well, stress is a scalar quantity, once the direction is fixed.
In general you have nine possible stresses, that can be grouped in a matrix ($F_x$, $F_y$ and $F_z$ in each plane $\textrm{XY}$, $\textrm{YZ}$ and $\textrm{XZ}$ respectively).
But I guess this is confusing you, so let's focus on only one of them: horizontal force on a cube's side.
Yes, force is a vector, and surface can be seen as a vector as well, but when you say "stress", it means "magnitude (scalar) of the force perpendicular to that element of area.
So the direction is previously fixed, and you only need a number to complete the information. That's what happens with pressure.
You can have a force on $y$ axis on the $\textrm{XY}$ plane, that's one stress $\sigma_{xy}$.
When you have force on $z$ axis over the $\textrm{XY}$ plane, that's force perpendicular to that plane. It is a scalar because it is "total force" (magnitude) over a surface whose orientation has already been described. That's why pressure is a scalar, and so are the other stresses.
If your surface does not coincide with those planes, you must take the projection of the force on each plane, and then compute the scalar stresses.
A: The state of stress at a given location within a material is described by the stress tensor, which is a 2nd order tensor.  The stress tensor helps us determine the force per unit area (called either the traction vector or the stress vector) on a surface of arbitrary orientation at the given location.  The traction vector on a surface is obtained from the stress tensor by applying the Cauchy Stress Relationship:  if $\mathbf{n}$ represents a unit normal vector to the surface (thus establishing its spatial orientation), the traction vector on the surface is obtained by taking the dot product of the stress tensor with the unit normal.  The traction vector thus obtained has both magnitude and direction, and, it is not necessarily perpendicular to the surface.  The component of the traction vector normal to the surface is called the normal stress; the component tangent to the surface is called the shear stress.
The isotropic part of the stress tensor is called the pressure (actually, minus the pressure), and, when dotted with the unit normal to any specified internal surface gives a traction vector of magnitude equal to -p, and direction normal to the surface; there is no shear stress component to the portion of the traction vector contributed by the (isotropic) pressure portion.
