@HTH: "Two vectors being multiplied will never give a scalar quantity"- with that all said we can assume dot product is not the same as 'multiplication' of two vectors but some other relation of them? I think I got confused because in this book it was written
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Yes. as I've tried to say multiple times. Thats why I don't like using the term multiplication, it confuses new people.
"Vectors can be multiplied in two ways, when they form a scalar product its called dot product and when they form a vector product its called a cross product"
That is the dumbest explanation ever.
"There are two 'multiplcation'-style operations that can be performed upon vectors. The first; Dot product, is calculated from two vectors A and B with the formula AxBx+AyBy=Dot product. The second, cross product, well to be quite frank I dont use a formula I use a x by x matrice and solve them quickly using determinants."
Either way your version is retardedly worded.
Applying even dimensional analyses the 'product' of two vectors lets say displacement*displacement should be displacement^2 or meter^2. I don't know what that'd be, a vector in two dimensions? Aren't all vectors in 2 dimensions with a component in each of them?
it is not vector*vectors. stop thinking it is. But yes a vector in a dimension D^a will have a component in each of the a directions. *2 in 2d, 3 in 3d, etc*
Secondly if no direction is involved in dot product then why does the formula go
A.B=ABcosx ??
Lol well frankly because that isnt the formula, that is the start to the proof of the communicative property of dot products. Where A.B = ABcos(theta) = BAcos(theta) = B.A
I can further simplify it using scalar projection or the geometric def'n of the dot product which utilizes a unit vector with an implicit magnitude of 1.. but id ratehr not explain it to you when you don't know the correct formulas, no offense..
I read all about the Bcosx is the projection of B over A. In real life example I can think of an object casting a shadow on the ground but that(the projection) would depend on the distance and position of light source too and who in their buttfucking right mind want to know the product of shadow and the area it covers on ground?
This part is quite simple lol. The scalar projection of A over B is equal to Acos(theta) because if you make a right triangle then A is the hypotenus and B is the adjacent, and cos=adj/hyp.
You can use it any time you need to find that length :p
I might be talking real nonsense here but i don't want to move on with the topic only rhyming
A.B=abcosx
AxB=absinx
without knowing what's actually going on.
EDIT: Oops forgot that you said it will be used in high level mathematics and I won't find an application of it at this level. How very comforting :-/
Right I'm gonna tell you right now that A.B might equal ABcosx but that is NOT the formula that you should use. That formula is used in proofs and shouldnt be taught at all at your level lol, it should have been a quick overview if anything just saying "yeah this proves A.B = B.A"
same for the cross product, thats the geometric definition, NOT the formula to find it.
A.B = Ax*Bx + Ay*By
or, the sum of the products of the components.
and the cross product is equal to... ah fuck it.
(x,y,z) X (x2,y2,z2) = (xc,yc,zc)
[ xc yc zc ]
[ x y z ]
[ x2 y2 z2 ]
where
xc = (y*z2) - (z*y2)
yc = (x*z2) - (z*x2)
zc = (x*y2) - (y-x2)
This can be remembered easily as taking the deteminant
[ a b ]
[ c d ]
det = ad-cb
in relation to the points that you wish to find, so for xc you disregard the row and column that it is in, same goes for yc, and zc.
Im sorry but outside of math you wont find an application for this any time soon, at least not a physical one. If you need any math help, again, ill usually respond to PM's. But I think you're either over thinking this or blessed with a very shitty teacher.