hi dudes, at the risk of this sounding like a totally unproductive question... i''m a noob at math so when i asked an engineer about the vector cross product he said the easiest way to find it on paper was to use determinants...blah blah. what i would like to know from the math gurus here...if possible is how it(vector cross product) came to be... some history if possible.i mean how did descartes derive his formula,if there is a proof could some one explain it in not so technical terms. just the other day i came to grips with why the distance formula is what it is.and i felt very satisfied with my puny little conquest.most of the time equations are just forced down my throat. i hope this is the place i can get a deeper and intelligble understanding of what they really are/mean. thank you sooo much for listening b

quote:when i asked an engineer about the vector cross product he said the easiest way to find it on paper was to use determinants...blah blah.

hmmn lets see.
Here is the definition of Vector Product aka cross product:
The vector product of two vectors a = [a1,a2,a3] and b = [b1,b2,b3] is a vector:
v = a x b
the new vector has the length:
|v| = |a| |b| sin @ where @ is the angle between A and B
and the direction of v is perpendicular to both a and b and such that a,b,v, in this order, form a right-hand triple.

In components, v = [v1,v2,v3] = a x b is
v1 = a2b3-a3b2, v2 = a3b1-a1b3, v3 = a1b2-a2b1

and that is a difinition. so to memorize this formula we use the determinants. as you may notice that:
v1 = DET [ a2 a3 ],[ b2 b3 ]
v2 = DET [ a3 a1 ],[ b3 b1 ]
v3 = DET [ a1 a2 ],[ b1 b2 ]
Hence v = v1 i + v2 j + v3 k is the expansion of the symbolical third-order determinant
a x b = DET ,[ a1 a2 a3 ],[ b1 b2 b3 ] </b> <br>by the first row. (we call it "symbolical} cecause the first row consists of vectors rather than numbers. Rememmber that in left-handed coordinates these determinants (2*) and (2**) must be multiplied by -1…<br><br>if you want deeper than that then you should refer to books. or if you want to know the general properties. The <i>vector product </i> is <b>not commutative </b> but <b>anticommutative </b> and that it is <b>not associative </b> …etc<br><br><br>hope this helps. and feel free to ask more…

hey dude,
thanks for the reply...though i''ll still need a lot of time to digest that...i gotta learn what determinants are now, damnn
really appreciate your help though
a question ...lets say that you didn''t know the formula
|v| = |a| |b| sin @ and you need to find the normal to two vectors in a plane.how would you have done such a thing ???
this way i can get a better idea of how descartes went about doing it and of course noticing a relationship and then equating it to |v| = |a| |b| sin @ .
again...before i get flamed with a napalm gun i am not a student looking to get his homework done...i''ve never done math beyound the tenth grade but the LOVE/DREAM of making games has made me learn programming and now the math.but math itself can drag you in a net of it''s own so this seams to be the ideal place to ask where math is game realted.
thanks again daher...do reply
b

the crossproduct can be generated with help of the dotproduct..

you want the normal of the plane?
you know one thing: there are an infinite amount of normals that fit. they all have the same direction (or inverse), but not the same length..


now, to define your normal, you need two direction-vectors in the plane (the ones you cross with eachothers to get the normal:D)
why you need that?
call them a and b..
you wan the normal n.

then you know
dot(n,a) == 0
dot(n,b) == 0

now you have 3 unknowns (n.x,n.y,n.z) and two equations. you get the idea:D

solve that.

other way.

take the three components, i,j,k. they define the 3 axis of our 3space ( i = (1,0,0), j = (0,1,0), k = (0,0,1) for clarification, but its not needed:D)

now, every vector is a sum of products with those 3 components:

v = ai + bj + cz (you can see thats the dotproduct with the vector (1,1,1) if you want..:D)

now, you can define a crossproduct by those rules:

i*i = 0
i*j = k
j*i = -k
etc..
those give you the normals you want, thats why you defined them that way (its just handy:D)

then just take your vectors again and use the same product on them

n = a*b
expands to
n = (a0i + a1j + a2k)*(b0i + b1j + b2k);

this is a simple product, you can solve it, and you get the crossproduct.

i don''t know wich one was when comming, but i learned the notation with i,j,k in my book first, and they let me calculate all sort of products that way very easily. define the rules you want on the simple base, then define your complex stuff with your simple base, and use the rules on your simple base again. a very handy way to solve big problems. determine what you really want to do. then find out where in your huge problem you find exactly that. do what you want on it, and you got it.

hope it inspires you a bit

"take a look around" - limp bizkit
www.google.com

It arose from quaternions. Hamilton discoved the quaterions in 1843, as an extension to the complex numbers and a tool for 3D rotations. It wasn''t until 1881 - 84 that Gibbs derived the vector cross product from the ''imaginary'' part of the product of ''imaginary'' quaternions.

For a while both systems (quaternions and vectors) were used, but there were disagreements over which was best and in the end Gibbs''s vector algebra won as the more easily understood and directly applicable system. Quaternions were largely forgotten until recently rediscovered as the best tool for 3D rotations.

The normal actualy is one of the applications of the Vector Product. Its main usage is for the Moment of a force and Velocity of a rotating body . it''s also widly used in Vector Differential Calculus .
But speeking of normals. every plane has one and only one normal vector. its length is one unit. The normal vector represents only a direction.
oy boy I am talking like an engineer again...

quote:Posted by boomji
...lets say that you didn''t know the formula
|v| = |a| |b| sin @ and you need to find the normal to two vectors in a plane.how would you have done such a thing ???
this way i can get a better idea of how descartes went about doing it and of course noticing a relationship and then equating it to |v| = |a| |b| sin @ .

This length is the area of the parallelogram with a and b as adjacent sides... hope this inspires you more.