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Deriving the algebraic definition the dot product 
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#1
Dec212, 04:47 AM

P: 333

Is there a way of deriving the algebraic definition of the dot product from the geometric definition without using the law of cosines?



#2
Dec212, 07:38 AM

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PF Gold
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What exactly do you mean by "the geometric definition"? The simplest "geometric" definition I know is that the dot product of vectors u and v is the product of the length of u and the length of the projection of v on u, but I suspect that you are thinking of "length of u times length of v times cosine of the angle between them". Since that has "cosine" already in it, if not using the "cosine law" per se, you certainly will need to use something that has a "cosine" in it.
From the definition "length of u times the length of the projection of v on u", I would start by setting up a coordinate system in which the positive xaxis lies along vector u. Then u has components <a, 0>. Taking <b, c> to be the components of v, the "projection of v on u" is just <b, 0> so the dot product is ab= ab+ c0, the usual component formula for the dot product in this case. Get the general formula by rotating axes. 


#3
Dec312, 11:27 AM

P: 333

Thanks HallsOfIvy, I'll try that.



#4
Dec312, 12:20 PM

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PF Gold
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Deriving the algebraic definition the dot product
Well, in the 2D case, given vectors "A", "B", with [itex](x_{a},y_{a})[/itex], [itex](x_{b},y_{b})[/itex], lengths a and b, respectively you may think in terms of rightangled triangles, and set up:
[tex]x_{a}=a\cos(\theta_{a})[/tex] [tex]y_{a}=a\sin(\theta_{a})[/tex] [tex]x_{b}=b\cos(\theta_{b})[/tex] [tex]x_{b}=b\sin(\theta_{b})[/tex] whereby we get: [tex]x_{a}x_{b}+y_{a}y_{b}=ab\cos(\theta_{a}\theta_{b})[/tex] 


#5
Dec412, 11:43 PM

P: 2

First define the dot product for A and B to be the product of their magnitudes and the cosine of the angle between them.
We can see geometrically that A.(B + C) = A.B + A.C (think about the component of B and C along A), and therefore (A + B).(C + D) = A.C + A.D + B.C + B.D. Choosing perpendicular axis, every vector can be written in terms of components, so A = a_1*i + a_2*j and B = b_1*i + b_2*j. Therefore A.B = a_1*b_1*i.i + a_2*b_2*j.j + a_1*b_2*i.j + a_2*b_1*j.i. Because i and j are unit vectors and perpendicular, i.i = 1, j.j = 1, i.j = 0, j.i = 0. So we are left with A.B = a_1*b_1 + a_2*b_2. 


#6
Dec512, 12:27 AM

P: 3,099

Also one thing to be aware of is that the algebraic defintion for vector dot and cross products only work when you have your vectors defined in a Euclidean space like our old favorite x,y,z or i,j,k.
and of course here's more info on it from wikipedia: http://en.wikipedia.org/wiki/Vector_dot_product 


#7
Dec512, 07:45 AM

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And the cross product is only defined for R^{3}, not general Euclidean spaces. 


#8
Dec512, 11:34 AM

P: 3,099

Its also interesting to note the beauty of the inner and outer products. The inner product is a projection of one vector on another and the outer product is the nonprojectable component of one vector on another.
Since the projection interpretation is valid for either vector then the resultant vector must be perpendicular to both and thus it becomes normal to the plane containing the two vectors. 


#9
Dec512, 05:53 PM

P: 333

Couldn't we just always work a coordinate system where the xaxis is parallel to the xcomponent of one vector?
Or is that what you mean when you said 'get the general formula by rotating axes?'. And would the formula only work in two dimensions? EDIT: Woops, didn't see the new posts! SO for rotating the axes, would it much more complicated in 3D? Thanks arildno, when you times the multiply the components and add them up, that just comes from the definition of dot product that HallsOfIvy mentioned right? ('length of u times the length of the projection of v on u'). 


#10
Dec812, 04:50 PM

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P: 1,722

Call these products the dot product of A and B. Either product divided by AB gives you the same number which depends only on the lines and not on the particular vectors. Call this number the cosine of the angle between the vectors. P.S. In Euclidean geometry the law of similar triangles which is what is used here is logically equivalent to the Pythagorean Theorem so using the Law of Cosines is really no different than what is posted here. 


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