The reason that there is a beam waist is just the fact that a Gaussian beam has certain properties:
1) These beams are a solution of the scalar Helmholtz wave equation. This implies that such beams contain a complex exponential, hence such beams will undergo damping at some point in space !
2) In the paraxial approximation, such beams give the same results as predicted by the Fresnel equations. This means that gaussian beams are a useful concept in optics.
3) Waves all exhibit the property of diffraction. This means that a beam cannot be focussed to one single point. This is actually the answer to your first question.
Now, if you look at the expression of a spherical Gaussian wave [tex]e^{ - \frac{x^2 + y^2}{w^2}}[/tex], the w defines the latitude of the spot (image). This w is calculated at that specific distance (starting from the position of the wave-source) where the wave's amplitude is reduced from it's original value to 1/e and the intensity is reduced to (1/e)². Now, if you take a plain gaussian wave (R = 0 and z = 0). You will see that this w reaches a minimal value which is called the waist. This is not just one single point because we are working with Gaussian waves here that exhibit the diffraction property !
To calculate the waist, you need to know at what angle the waves diverges starting from the waist. This angle theta is equal to [tex]\theta = \frac{ \lambda}{ \pi w'}[/tex]. The w' is the magnitude of the waist. This answers your second question.
More generally :
If you know the curvature radius R of the beam at a certain position and the spot magnitude w1 than you can calculate both the position and the magnitude of the waist with these formula's:
position
[tex]z = \frac {R}{1 + ( \frac{\lambda R}{\pi w1^2})^2}[/tex]
magnitude
[tex]w' = \frac {w1}{ \sqrt{1 + ( \frac{\pi w1^2}{\lambda R})^2}}[/tex]
marlon