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dot product and it's derivation

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hi, googled around first but did not find what i was looking for. any way,could any one tell me why is the dot product ||A||*||B||cos theta. Every where i read up this was a given(cos theta). now i know i may not yet be abel to digest the proof even if it was explained but i''m hoping some one could post a visual explanation or point me to one...please. thanks for your time. b

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The definition of the dot product for two vectors and is Ax*Bx + Ay*By. Also, Aθ is arctan(Ay/Ax) and Bθ is arctan(By/Bx). SO:

dot(A,B) = Ax*Bx + Ay*By
= ||A||cos(Aθ ) * ||B||cos(Bθ ) + ||A||sin(Aθ ) * ||B||sin(Bθ )

= ||A|| * ||B|| * cos(Aθ ) * cos(Bθ ) + ||A|| * ||B|| * sin(Aθ ) * sin(Bθ )

= (||A|| * ||B||)(cos(Aθ ) * cos(Bθ ) + sin(Aθ ) * sin(Bθ ))

and, by some product-to-sum trig identities:

= (||A|| * ||B||)((1/2 * cos(Aθ - Bθ ) + cos(Aθ + Bθ ))+(1/2 * cos(Aθ - Bθ ) - cos(Aθ + Bθ )))

= (||A|| * ||B||) * 1/2 * (cos(Aθ - Bθ ) + cos(Aθ + Bθ )+cos(Aθ - Bθ ) - cos(Aθ + Bθ ))

= (||A|| * ||B||) * 1/2 * (cos(Aθ - Bθ )+cos(Aθ - Bθ ))

= (||A|| * ||B||) * cos(Aθ - Bθ )

How appropriate. You fight like a cow.

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Let''s do it backwards.

Let''s say you have 2 vectors in 3D, and you want to find |A|*|B|*cos theta. Does this reduce to Ax * Bx + Ay * By + Az * Bz? Let''s see.

|B|*cos theta is the length of the projection of vector B onto vector A. Let''s find the point on vector A of this projection.

Let''s call the coordinates of this point (Cx, Cy, Cz).

Now, let''s create a plane that has its normal as vector A - and let''s set this plane to intersect with the endpoint of vector B. The intersection of this plane with vector A will yield (Cx, Cy, Cz)!

This plane is:
Ax * x + Ay * y + Az * z = K, where K is some constant

Now, this plane must intersect with (Bx, By, Bz) as stated above.

Ax * Bx + Ay * By + Az * Bz = K

So the equation of the plane is:

Ax * x + Ay * y + Az * z = Ax * Bx + Ay * By + Az * Bz

Let''s find out how far this plane is along vector A.

Let''s represent vector A as a parametric line:

x = Ax * t / |A|
y = Ay * t / |A|
z = Az * t / |A|

We must divide by |A| in order to make t the actual distance along vector A. (i.e. an increment of 1 in t travels 1 unit)

Sub that in to the plane:

Ax * Ax * t / |A| + Ay * Ay * t / |A| + Az * Az * t / |A| = Ax * Bx + Ay * By + Az * Bz

Ax^2 * t / |A| + Ay^2 * t / |A| + Az^2 * t / |A| = Ax * Bx + Ay * By + Az * Bz

Ax^2 * t + Ay^2 * t + Az^2 * t = |A| * (Ax * Bx + Ay * By + Az * Bz)

Since Ax^2 + Ay^2 + Az^2 = |A|^2:

|A|^2 * t = |A| * (Ax * Bx + Ay * By + Az * Bz)

t = (Ax * Bx + Ay * By + Az * Bz)/|A|

Since t is the distance along vector A of the projection of vector B, t = |B| * cos theta

Therefore:

|A| * |B| * cos theta

= |A| * t

= |A| * (Ax * Bx + Ay * By + Az * Bz)/|A|

= Ax * Bx + Ay * By + Az * Bz

Wow ... that actually worked!

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oh lol... you wanted a visual explanation.

Well, have fun with that crazy stuff I wrote.

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Take A = |A|*a, B = |B|*b so that the vectors a=(ax,ay) and b=(bx,by) are unit vectors.

1) I think the factors |A| and |B| are clear then (put it into the equations if you don´t believe it, also the scalar product is bilinear by definition).

2) a*b is the projection of vector a on vector b (which happenes to be the same as the projection of b on a). Let both a and b lie in the xy-plane and let b be a unit vector in the x-direction (you might want to think about why/if that assumption is ok in a general case after you understood the following).
Draw both vectors in your coordinate system. Draw a triangle with points (0,0), (ax, ay), (ax, 0) which is a recangular triangle (vector (ax,0) is the projection of a on b as you will see).
By the geometrical definition of cos you´ll see that cos(theta) = |(ax,0)|/|(ax,ay)| = |projection(a -> b)|/1 = a*b.

EDIT: The |(ax,0)| above is not completely correct since it can also be negative when a is in the 2nd or 3rd quadrant but you´ll see from your picture what I meant.

[edited by - Atheist on October 22, 2003 2:56:34 AM]

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Actually guys, the dot product between 2 vectors A, B is DEFINED as |A||B|cos theta. Most people (mistakingly) get it the wrong way round, A.xB.x + A.yB.y + A.zB.z is NOT the definition of the dot product, it is merely a reduction of the definition into component form. The reduction is just an easier way of computing the dot product, and it follows from several different methods of reduction: one of which is Tron3k''s reduction.

Here is another reduction which I found interesting and in my opinion more visually and mathematically intuitive.

(1) A.B = |A||B|cos(theta) (by defn of dot product)

If you think of the length of the 3 vectors |A|,|B| and |B-A| as the lengths of the sides of a triangle, you can apply the law of cosines here too (boomji, to visualize this, draw the 2 vectors A and B onto a graph, now the vector from A to B will be given by B-A. The triangle formed by these 3 vectors is applied to the law of cosines for a triangle :D)
c^2 = a^2 + b^2 - 2.a.b.cos(C)
In this case, we substitute: |B-A| for c, |A| for a, |B| for b
and we obtain:

(2) |B-A|^2 = |A|^2 + |B|^2 - 2 |A|.|B|.cos(Theta) (by law of cosines)

Remember now, that Theta is the angle between the 2 vectors A, B.
Notice the common term |A||B|cos(Theta) in both equations. We now equate equation (1) and (2), and obtain

A.B = (-(|B-A|^2) + |A|^2 + |B|^2) / 2
and hence
A.B = (-((B.x^2 - 2A.xB.x + A.x^2) + (B.y^2 - 2A.yB.y + A.y^2) + (B.z^2 - 2A.zB.z + A.z^2)) + A.x^2 + A.y^2 + A.z^2 + B.x^2 + B.y^2 + B.z^2) / 2
(by pythagorean length of a vector) and thus
A.B = A.xB.x + A.yB.y + A.zB.z

This one I think is more visually explainable. Unfortunately, I don''t have the time to draw a diagram, sorry!

Hope this helped!

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Where did you get that way of definition (and what makes you think your source is THE definition)? I always thought the dot product is a special scalar product which is (in the form I know) given as a multilinear positive definite symmetric form over a vector space.
Also if the two forms are equal (which is only true in the case of a real vector space because you can´t define an angle in general) where is the point in marking one form as THE definition?

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hi there,
thanks so much folks...and pretty detailed as well.
the cosine rule and it's involvement here has made most of the cogs in my block head turn so i will definately look into that area more(did not do math beyound 10th grade...came right back at me and bit me in the moon there didn't it ).
thanks a ton guys...going to save this page and thoroughly go through it.

b

[edited by - boomji on October 22, 2003 1:51:18 PM]

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Hey Athiest, don''t be mistaken - the idea that that is the definition of dot-product is not just simply an idea that popped into my head. It was an idea strictly enforced by the mathematics academics whilst I was doing my degree. More generally, though the definition pertains to Euclidean space rather than Inner-product spaces.Sorry to have made such a broad statement - I guess that''s just asking for trouble (even if I am right :D) ... but by all means, prove me wrong

The point of proper definitions: learning and common language, although in this case they may clear up confusion of certain concepts. Such as the origin of the projection of a vector onto another vector, which has it''s basis in geometric and trigonometric methods and which is more easily explained by the definition |A||B|cosC of the dot product.

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