It is defined as:
Work is the dot product of force and displacement.
From these it follows directly that the dot product of two vectors a = [a1 a2 a3] and b = [b1 b2 b3] given in coordinates can be computed particularly easily:
Properties
The definition has the following consequences:
or, using matrix multiplication and treating the vectors as 1-by-3 matrices:
The dot product satisfies all the axioms of an inner product. In an abstract vector space, the notion of angle between the elements of the space can be defined in terms of the inner product.
;That is, given :a·b = a1b1 + a2b2 + a3b3
;derive :a·b = |a|.|b|.cos(θ),
;or using nomenclature x for |x| :a·b = a.b.cos(θ).
This proof is shown for 3-dimensional vectors, but is readily extendable to N-dimensional vectors given mutually perpendicular unit vectors.
;Consider a vector :v =v1i + v2j + v3k.
;Repeated application of the Pythagorean theorem determines that :v2 =(v12 + v22 + v32).
;which is the same formula for the dot product of the vector v with itself, thus :v·v = v2.
;Now consider two vectors a and b from the origin and separated by an angle θ. A third vector c may be defined as :c = a-b.
;Using the law of cosines, we have :c2 = a2 + b2 - 2.a.b.cos(θ).
;And substituting the dot product for the squared lengths, we get :c·c = a·a + b·b - 2.a.b.cos(θ).
;But as c = a - b, we also have : c·c = (a - b)·(a - b).
;which expands to :c·c = a·a + b·b - 2.a·b.
;Then merging the two c·c equations we obtain :a·a + b·b - 2.a·b = a·a + b·b - 2.a.b.cos(θ).
;Subtracting a·a + b·b from both sides leaves :- 2.a·b = - 2.a.b.cos(θ).
;And dividing by -2 derives the final :a·b = a.b.cos(θ).
See also: Cross productProof that the two forms of definition are equivalent