Figure 1: {\bf a} \cdot{\bf b} = ab\cos\theta.
If the vectors a and b have magnitudes a and b respectively, and if the angle between them is \theta, then the scalar product of a and b is defined to be
{\bf a} \cdot{\bf b} = ab\cos\theta.
The scalar product is quite clearly commutative: $${\bf a} \cdot{\bf b} = {\bf b}\cdot{\bf a}.$$
It's less easy to show that it's also distributive over addition: $${\bf a} \cdot({\bf b}+{\bf c})={\bf a}\cdot{\bf b}+{\bf a}\cdot{\bf c}.$$ However, this is indeed the case, and it's why we call it a "product" at all.
Note that
From these results, together with the distributivity property, it's easy to show that the scalar product of $$a = a_1{\bf i}+a_2{\bf j}+a_3{\bf k}$$ and $$b = b_1{\bf i}+b_2{\bf j}+b_3{\bf k}$$ is $$a \cdot{\bf b}=a_1b_1+a_2b_2+a_3b_3.$$ This yields an easy method for calculating the angle between two vectors given in component form. For example, the angle \theta between the vectors $$a = 9{\bf i} - 2{\bf j} -6{\bf k}$$ and $$b = {\bf i} - 2{\bf j} + 2{\bf k}$$ is calculated as follows. We know that $$a\cdot{\bf b}=ab\cos\theta.$$ That is, $$1\times9+(-2)\times(-2)+(-6)\times2=\sqrt{9^2+2^2+6^2}\times\sqrt{1^2+2^2+2^2}\cos\theta,$$ or simply $$1=33\cos\theta.$$ it follows that \theta = \cos^{-1}(1/33) \approx 1.54 radians.
Figure 2: a\cos\theta. , the length of the projection of a in the direction of b
Note that the dot product of a with a unit vector \hat {\bf b} is a \cos \theta, the length of the projection of a in the direction of b.