structural:scholar_edition, structural:expansion, quality:corruption, linguistic:voice_change
Equivalence of the definitions

If e 1, ⋯, e n {\displaystyle \mathbf {e} _{1},\cdots,\mathbf {e} _{n}} are the standard basis vectors in R n {\displaystyle \mathbf {R} ^{n}}, then we may write a = = ∑ i a i e i b = = ∑ i b i e i. {\displaystyle {\begin{aligned}\mathbf {a} &==\sum _{i}a_{i}\mathbf {e} _{i}\\\mathbf {b} &==\sum _{i}b_{i}\mathbf {e} _{i}.\end{aligned}}} The vectors e i {\displaystyle \mathbf {e} _{i}} 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, e i ⋅ e i = 1 {\displaystyle \mathbf {e} _{i}\cdot \mathbf {e} _{i}=1} and since they form right angles with each other, if i ≠ j {\displaystyle i\neq j}, e i ⋅ e j = 0. {\displaystyle \mathbf {e} _{i}\cdot \mathbf {e} _{j}=0.} Thus in general, we can say that: e i ⋅ e j = δ i j, {\displaystyle \mathbf {e} _{i}\cdot \mathbf {e} _{j}=\delta _{ij},} where δ i j {\displaystyle \delta _{ij}} is the Kronecker delta.

Also, by the geometric definition, for any vector e i {\displaystyle \mathbf {e} _{i}} and a vector a {\displaystyle \mathbf {a} }, we note that a ⋅ e i = ‖ a ‖ ‖ e i ‖ cos θ i = ‖ a ‖ cos θ i = a i, {\displaystyle \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},} where a i {\displaystyle a_{i}} is the component of vector a {\displaystyle \mathbf {a} } in the direction of e i {\displaystyle \mathbf {e} _{i}}. 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 a ⋅ b = a ⋅ ∑ i b i e i = ∑ i b i (a ⋅ e i) = ∑ i b i a i = ∑ i a i b i, {\displaystyle \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},} which is precisely the algebraic definition of the dot product. So the geometric dot product equals the algebraic dot product.