-that correpsonds to the length of vector \vec{a}.
-Evaluating the scalar product $(\vec{a},\vec{b})$ by the sum representation of \eqref{eq:vec_sum} \ldots
+that just corresponds to the length of vector $\vec{a}$.
+Evaluating the scalar product $(\vec{a},\vec{b})$ by the sum representation of \eqref{eq:vec_sum} leads to
+\begin{equation}
+(\vec{a},\vec{b})=(\sum_i\vec{e}_ia_i,\sum_j\vec{e}_jb_j)=
+\sum_i\sum_j(\vec{e}_i,\vec{e}_j)a_ib_j \text { ,}
+\end{equation}
+which is equal to \eqref{eq:vec_sp} only if
+\begin{equation}
+(\vec{e}_i,\vec{e}_j)=
+\delta_{ij} = \left\{ \begin{array}{lll}
+0 & {\rm for} ~i \neq j \\
+1 & {\rm for} ~i = j \end{array} \right.
+\text{ (Kronecker delta symbol),}
+\end{equation}
+i.e.\ the basis vectors are mutually perpendicular (orthogonal) and have unit length (normalized).
+Such a basis set is called orthonormal.
+The component of a vector can be obtained by taking the scalar product with the respective basis vector.
+\begin{equation}
+(\vec{e}_j,\vec{a})=(\vec{e}_j,\sum_i \vec{e}_ia_i)=
+\sum_i (\vec{e}_j,\vec{e}_i)a_i=
+\sum_i\delta_{ij}a_i=a_j
+\end{equation}
+Inserting the expression for the coefficients into \eqref{eq:vec_sum}, the vector can be written as