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Dot Product Derviation

By Michael Henderson
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The dot product or inner product in Euclidean Space $A\cdot B$ has two definitions:

  1. Algebraically defined as: $$A \cdot B = \sum_{i=1}^{n}A_i \cdot B_i=A_1B_1 + A_2B_2 ... A_nB_n$$

  2. Geometrically defined as: $$A \cdot B = \Vert A \Vert \Vert B \Vert \cos{\theta}$$

Is there a way to prove that they are equal other than simply saying they are defined so?

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1 Answer

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Note that if $A$ and $B$ are vectors representing two sides of a triangle, then the third side is represented by vector $C=B-A$. Let the corresponding magnitudes of the sides be $a,b,c$

The cosine rule tells us that $c^2=a^2+b^2-2ab \cos \theta$

Computing the squares of the magnitudes from the components (using Pythagoras) we obtain $$2ab\cos \theta=\sum A_i^2+\sum B_i^2-\sum (B_i-A_i)^2=2\sum A_iB_i$$and the result follows.


Really this is why the scalar product is a useful definition - if it didn't have this geometric aspect we'd be using a different idea. For example, this captures the idea of the component of a force acting in a given direction.

The scalar product has the advantage of expressing the geometric reality without having to specify a basis.

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