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- thomas5267

I do not believe cross product is even defined in \(\mathbb{R}^n,\,n\neq3\).

- inkyvoyd

Cross product is definitely defined in R^2. Not sure about R^4.
Also I guess I done messed up
http://math.stackexchange.com/questions/720813/do-four-dimensional-vectors-have-a-cross-product-property
but what about the dot product?

- IrishBoy123

@empty @Michele_Laino

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## More answers

- thomas5267

Scroll down on the webpage you provided. The answer clearly said that cross product does not exist outside of \(\mathbb{R}^3\) and \(\mathbb{R}^7\).
Furthermore, dot product and cross product are entirely different even on \(\mathbb{R}^3\)! Dot product of two vectors is a scalar and cross product of two vector is a vector.

- thomas5267

https://math.stackexchange.com/questions/424482/cross-product-in-mathbb-rn

- inkyvoyd

@thomas5267 , that was entirely my point - why else would i provide a webpage clearly stating that the cross product doesn't exist outside of those contexts? Anyways, I am familiar with the cross product and dot product - I am searching for proofs, in particular, the one relating the algebraic and geometric properties of the dot product in R^n. I have had trouble finding that online.

- zzr0ck3r

relate?? one is a vector and one is a scalar

- zzr0ck3r

They are entirely different operations

- inkyvoyd

@zzr0ck3r , I mean this.https://en.wikipedia.org/wiki/Dot_product#Algebraic_definition

- thomas5267

Oh! I get what you are asking now.
You are not asking for algebraic equivalence of the dot and cross products in \(\mathbb{R}^n\) and the geometric equivalence of the dot and cross products in \(\mathbb{R}^n\).
You are asking the algebraic and geometric equivalence of the dot product in \(\mathbb{R}^n\) and algebraic and geometric equivalence of the cross product in \(\mathbb{R}^n\)?

- inkyvoyd

Yes, the latter of which does not really exist as n can only equal 3 and 7 which you pointed out to me. Apologies for my misplaced modifier in the question as I did not get much sleep and did not even realize it was being interpreted as such.

- thomas5267

The lack of sleep can do wonderful things to humans indeed!

- zzr0ck3r

ahh, I see.

- thomas5267

The wikipedia link on dot product kind of answers your question.
https://en.wikipedia.org/wiki/Dot_product#Definition

- inkyvoyd

you mean "In modern presentations of Euclidean geometry, the points of space are defined in terms of their Cartesian coordinates, and Euclidean space itself is commonly identified with the real coordinate space Rn. In such a presentation, the notions of length and angles are not primitive. They are defined by means of the dot product: the length of a vector is defined as the square root of the dot product of the vector by itself, and the cosine of the (non oriented) angle of two vectors of length one is defined as their dot product. So the equivalence of the two definitions of the dot product is a part of the equivalence of the classical and the modern formulations of Euclidean geometry."?

- thomas5267

Scroll down a little bit. There is a section called "Algebraic definition" and a section called "Geometric definition". Scroll down a little bit more and there is a section called "Equivalence of the definitions".

- inkyvoyd

phooey. I assume I won't be able to get away with understanding this guy without having first taken linear algebra?

- zzr0ck3r

lol good idea :)

- thomas5267

I believe there is a proof on the algebraic and geometric equivalence of the dot product on \(\mathbb{R}^2\) using cosine rule and vector addition. Is \(\mathbb{R}^2\) all you are after?

- thomas5267

\[
\mathbf{a}=(a_1,a_2)\\
\mathbf{b}=(b_1,b_2)\\
\mathbf{c}=\mathbf{b}-\mathbf{a}=(b_1-a_1,b_2-a_2)\\
\|\mathbf{c}\|^2=\|\mathbf{a}\|^2+\|\mathbf{b}\|^2-2\|\mathbf{a}\|\|\mathbf{b}\|\cos(\theta)\\
(b_1-a_1)^2+(b_2-a_2)^2=\left(a_1^2+a_2^2\right)+\left(b_1^2+b_2^2\right)-2\|\mathbf{a}\|\|\mathbf{b}\|\cos(\theta)\\
b_1^2-2a_1b_1+a_1^2+b_2^2-2a_2b_2+b_2^2=\left(a_1^2+a_2^2\right)+\left(b_1^2+b_2^2\right)-2\|\mathbf{a}\|\|\mathbf{b}\|\cos(\theta)\\
-2a_1b_1-2a_2b_2=-2\|\mathbf{a}\|\|\mathbf{b}\|\cos(\theta)\\
a_1b_1+a_2b_2=\|\mathbf{a}\|\|\mathbf{b}\|\cos(\theta)\\
\mathbf{a}.\mathbf{b}=\|\mathbf{a}\|\|\mathbf{b}\|\cos(\theta)
\]
I have no idea what I did but this is the proof.

- inkyvoyd

hmm.. I am familiar with that proof (thank you for typing it out), but I was wondering if you could extend it to R^3 or ^n...

- beginnersmind

Think about how difficult it is to find angles geometrically in 3 dimensions. If you already have coordinates the best way to do it is to calculate the dot product using coordinates and work backward to find the angle.
The purely geometric way would be to find the plane determined by the two vectors and work is '2 dimensions'.
The general proof would look something like this (works for R^3 or R^n as well):
start with vectors
u(u_1,u_2,...,u_n) and v(v_1, v_2, ... ,v_n)
Find the plane determined by u and v, and define a coordinate system (x,y) on that plane, with perpendicular axes and unit coordinates. Prove that if
u(x_u, y_u) and v(x_v, y_v) are the coordinates of u and v in this new coordinate system, then x_u*x_v + x_v*y_v gives the same result as the dot product u*v calculated using the n coordinates. This is the hard part, and impossible without a fair bit of linear algebra.
Now you can use the two dimensional proof to prove that the value calculated using lengths and angles is the same as the value calculated using 2 coordinates, which is the same as the value calculated using n coordinates.

- thomas5267

What exactly is u(u_1,u_2,...,u_n) and v(v_1, v_2, ... ,v_n) ? Are they vectors or what?

- beginnersmind

u and v are vectors in R^n. u_1, u_2, ... u_n are their coordinates in the standard basis.

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