Editing Dot product (section)

=== Equivalence of the definitions ===
If <math>\mathbf{e}_1,\cdots,\mathbf{e}_n</math> are the [[standard basis|standard basis vectors]] in <math>\mathbf{R}^n</math>, then we may write
<math display="block">\begin{align}
\mathbf a &= [a_1 , \dots , a_n] = \sum_i a_i \mathbf e_i \\
\mathbf b &= [b_1 , \dots , b_n] = \sum_i b_i \mathbf e_i.
\end{align}
</math>
The vectors <math>\mathbf{e}_i</math> are an [[orthonormal basis]], which means that they have unit length and are at right angles to each other.  Since these vectors have unit length,
<math display="block"> \mathbf e_i \cdot \mathbf e_i = 1 </math>
and since they form right angles with each other, if <math>i\neq j</math>,
<math display="block"> \mathbf e_i \cdot \mathbf e_j = 0 .</math>
Thus in general, we can say that:
<math display="block"> \mathbf e_i \cdot \mathbf e_j = \delta_ {ij} ,</math>
where <math>\delta_{ij}</math> is the [[Kronecker delta]].

[[File:Skalarprodukt geometrisch.svg|thumb|upright=1.0|Vector components in an orthonormal basis]]

Also, by the geometric definition, for any vector <math>\mathbf{e}_i</math> and a vector <math>\mathbf{a}</math>, we note that
<math display="block"> \mathbf a \cdot \mathbf e_i = \left\| \mathbf a \right\| \left\| \mathbf e_i \right\| \cos \theta_i = \left\| \mathbf a \right\| \cos \theta_i = a_i ,</math>
where <math>a_i</math> is the component of vector <math>\mathbf{a}</math> in the direction of <math>\mathbf{e}_i</math>. The last step in the equality can be seen from the figure.

Now applying the distributivity of the geometric version of the dot product gives
<math display="block"> \mathbf a \cdot \mathbf b = \mathbf a \cdot \sum_i b_i \mathbf e_i = \sum_i b_i ( \mathbf a \cdot \mathbf e_i ) = \sum_i b_i a_i= \sum_i a_i b_i ,</math>
which is precisely the algebraic definition of the dot product.  So the geometric dot product equals the algebraic dot product.