How to find cross product of 2 vectors in R^4 ?

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How to find cross product of 2 vectors in R^4 ?

Mathematics
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\(\left[\begin{matrix} 1\\1\\0\\0\end{matrix}\right]\) and \(\left[\begin{matrix} -1\\0\\1\\0\end{matrix}\right]\)
Let v is the required vector, so that v orthogonal to both vector above. It is easy to see that \(v= \left[\begin{matrix} 0\\0\\0\\1\end{matrix}\right]\) satisfies the condition. But I don't want it. I want the method to find it out, not just trial and error. Please, help me out
as far as i can recollect, cross product is only defined in R^3 but I do recall a paper ina journal that discussed the usefulness of a nxm determinant where n /= m

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So, how can we find it out in this case?
we can use dot product also. right? Let v = , hence v . v1 = < a, b, c, d> .< 1,1,0,0> = 0 (we need it) and v. v2 = < a, b, c, d> .< -1,0,1,0> =0 also. But it is hard to find out the result
yes, dotproduct will determine if a vector is parralel to another.
for the first one, it is a + b =0 the second one , it is -a + c = 0 hence a = -b, and c, d are arbitrary, a = c and b, d are arbitrary then v can be < 1, -1, 1, 0> , but it doesn't satisfy v.v1 =0 or v.v2 =0 :(
the plane in R^4 is most likely teh sae approach as a circle to a sphere, add on the missing dimension
I know my logic is flaw at somewhere, but don't know how to fix yet
n = (a,b,c,d) a(x-xo) + b(y-yo) + c(z-zo) + d(ko) = 0 but then im thinking that in R^4 there are multiple 'normals' to a plane
k-ko that is
oh, i got it. the first one gives me since I can pick any of c, d the second one gives me since i can pick any of b and d But they must be the same. that is < a, -a, 1, 1> =< a, 1, a, 1> hence a = a, -a =1 , 1 = a. it is possible if and only if a =0, hence the required v is v =< 0,0,0,1> Am I right?
abcd 1100 ----- a+b = 0 abcd -1010 ------ -a+c= 0 a = a b = -a c = a d = d
|dw:1444571517011:dw|
1,-1,1,2 ; and -1,1,-1,7 1-112 1 100 ----- 1-100 = 0 1-112 -1010 ------ -1010 = 0 -11-17 1 1 00 ------ -1100 = 0 1-100 = 0 -11-17 -10 10 ------ 10-10 = 0
but dotproduct is not cross product ... so i think your attempts are misguided
The goal of cross product is to find out a vector perpendicular to the given vectors, right?
Hence if we can't find it out by cross while dot can help us, why not?
that is an application of it yes, but i am not sure if that is the only reason for it.
Do you know something about flat ?
I read a paper but I don't get what it is.
'flat' doest ring any bells ...
ok, thanks for the help. :)
i think the concept of trying to find orthogonal vectors in R^4 given a R^2 object; is similar to finding them in R^3 for and R^1 object. given the vector (1,2,3) find all the vectors perpendicular to it. well, this one vector is the normal to an infinite number of vectors that form a plane. |dw:1444572269541:dw|
there are going to be an infinite number of vectors in R^4 that are perpendicular to a R^2 object.
I have so many problems like this but I don't get it. ha!! find a vector v such that v . < -3,1,1,0,1 > = v .< -1,1,0,1,0> =0
|dw:1444572525474:dw| the 'normals' in R^4 to an R^2 plane can be thought of to form a 'cone' that is perpendicular to the plane itself and its normal ...
find "a" vector that mets the condition it amounts to a system of n equations, in more than n unknowns.
a b c d e -3,1,1,0,1 ----------- -3a+b+c+e = 0 a b c d e -1,1,0,1,0 ------------ -a+b+d = 0
define value for cde, 1,2,3 then we reduce this to a 2 system in ab -3a+b+1+3 = 0 -a+b+2 = 0 determine (a,b,1,2,3) as your vector
or (a,b,0,0,0) or .... whatever is simplest for you to deduce
just a thought, but it may be related to finding pivot points for those nxm matrix setup to evaluate them
-1,1,0,1,0 -3,1,1,0,1 1, -1, 0,-1, 0 -1,-1/3,-1/3, 0,-1/3 -------------------- 0,-4/3,-1/3,-1,-1/3 0, 1,1/4,3/4, 1/4 1,-1, 0, -1, 0 0 , 1 ,1/4, 3/4, 1/4 1, 0,-1/4,-1/4,-1/4 0 , 1, 1/4, 3/4, 1/4 a = (c+d+e)/4 b = -(c+3d+e)/4 just practicing my matrix stuff at the moment is all :)
I will. Thank you so much. :)
|dw:1444576604879:dw|
|dw:1444580267634:dw|

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