A community for students.
Here's the question you clicked on:
 0 viewing
Loser66
 one year ago
How to find cross product of 2 vectors in R^4 ?
Loser66
 one year ago
How to find cross product of 2 vectors in R^4 ?

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\(\left[\begin{matrix} 1\\1\\0\\0\end{matrix}\right]\) and \(\left[\begin{matrix} 1\\0\\1\\0\end{matrix}\right]\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Let 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2as 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1So, how can we find it out in this case?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1we can use dot product also. right? Let v =<a, b, c, d> , 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2yes, dotproduct will determine if a vector is parralel to another.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1for 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 :(

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2the plane in R^4 is most likely teh sae approach as a circle to a sphere, add on the missing dimension

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I know my logic is flaw at somewhere, but don't know how to fix yet

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2n = (a,b,c,d) a(xxo) + b(yyo) + c(zzo) + d(ko) = 0 but then im thinking that in R^4 there are multiple 'normals' to a plane

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, i got it. the first one gives me <a, a, 1, 1> since I can pick any of c, d the second one gives me <a, 1, a , 1> 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?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2abcd 1100  a+b = 0 abcd 1010  a+c= 0 a = a b = a c = a d = d

amistre64
 one year ago
Best ResponseYou've already chosen the best response.21,1,1,2 ; and 1,1,1,7 1112 1 100  1100 = 0 1112 1010  1010 = 0 1117 1 1 00  1100 = 0 1100 = 0 1117 10 10  1010 = 0

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2but dotproduct is not cross product ... so i think your attempts are misguided

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1The goal of cross product is to find out a vector perpendicular to the given vectors, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hence if we can't find it out by cross while dot can help us, why not?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2that is an application of it yes, but i am not sure if that is the only reason for it.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Do you know something about flat ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I read a paper but I don't get what it is.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2'flat' doest ring any bells ...

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1ok, thanks for the help. :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2i 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2there are going to be an infinite number of vectors in R^4 that are perpendicular to a R^2 object.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2dw: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 ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2find "a" vector that mets the condition it amounts to a system of n equations, in more than n unknowns.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2a 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2define 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2or (a,b,0,0,0) or .... whatever is simplest for you to deduce

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2just a thought, but it may be related to finding pivot points for those nxm matrix setup to evaluate them

amistre64
 one year ago
Best ResponseYou've already chosen the best response.21,1,0,1,0 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 :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I will. Thank you so much. :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.