Question 1087143
You should revise your question.  

——
You have provided the geometric definition for the dot product (aka inner product) of two vectors.  The definition can be accepted straight away.    Are you asking to prove the algebraic definition of the dot product is equal to the geometric definition of the dot product?  

Algebraic:
{{{ a_dot_b = a[1]*b[1] + a[2]*b[2] + a[3]*b[3] }}} + … + {{{ a[n]*b[n] }}}

Geometric:
{{{ a_dot_b = abs(a)abs(b)*cos(theta) }}}   where {{{theta}}} = angle between the two vectors


As I thought about it, I figured that must be what you want.  Here is the proof:
I will use capital letters for vectors, lower case for scalers.   
Also |A| means the magnitude of A, found using ordinary root-of-squares method.

Let {{{ E[i] }}} be a set of unit vectors in the n-dimensional space.
So  
{{{ E[i]*E[j] = 0  }}} if {{{  i<>j }}}
 {{{  E[i]*E[j] = 1 }}} if {{{   i=j ) }}}

<pre>
{{{ A = sum(a[i]E[i], i=1, n ) }}}    (1)     and
{{{ B = sum(b[i]E[i], i=1, n) }}}     (2)    <—<<<  Remember {{{ a[i] }}} and {{{ b[i] }}} are scalers 
</pre>

and the algebraic dot product is defined as:
{{{ A*B = sum(a[i]b[i], i=1, n) }}}     (*)
—

Now, use the geometric dot product definition to find {{{ A*E[i] }}} : 
{{{ A*E[i] = abs(A)abs(E[i])cos(theta) = abs(A)*1*cos(theta) = a[i] }}}  <—<<< Same {{{ a[i] }}} as in (1)
(and similarly for {{{ B*E[i] }}} )
{{{ a[i] }}} and {{{ b[i] }}} can be thought of as the magnitude of the component of A, B that project in the {{{E[i]}}} direction, respectively.     

and substituting for B:
{{{ A*B = A*sum(b[i]E[i], i=1, n) }}}
Finally, using the distributive property of dot products  {{{ C*kD = kC*D }}} :
{{{ A*B = A*sum(b[i]E[i], i=1, n)  = sum(b[i]A*E[i], i=1, n)  = highlight(sum(b[i]a[i], i=1, n)) }}}  <—<<< Same as (*)