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Loser66

  • one year ago

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

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  1. Loser66
    • one year ago
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    \(\left[\begin{matrix} 1\\1\\0\\0\end{matrix}\right]\) and \(\left[\begin{matrix} -1\\0\\1\\0\end{matrix}\right]\)

  2. Loser66
    • one year ago
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    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

  3. amistre64
    • one year ago
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    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

  4. Loser66
    • one year ago
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    So, how can we find it out in this case?

  5. Loser66
    • one year ago
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    we 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

  6. amistre64
    • one year ago
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    yes, dotproduct will determine if a vector is parralel to another.

  7. Loser66
    • one year ago
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    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 :(

  8. amistre64
    • one year ago
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    the plane in R^4 is most likely teh sae approach as a circle to a sphere, add on the missing dimension

  9. Loser66
    • one year ago
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    I know my logic is flaw at somewhere, but don't know how to fix yet

  10. amistre64
    • one year ago
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    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

  11. amistre64
    • one year ago
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    k-ko that is

  12. Loser66
    • one year ago
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    oh, 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?

  13. amistre64
    • one year ago
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    abcd 1100 ----- a+b = 0 abcd -1010 ------ -a+c= 0 a = a b = -a c = a d = d

  14. Loser66
    • one year ago
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    |dw:1444571517011:dw|

  15. amistre64
    • one year ago
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    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

  16. amistre64
    • one year ago
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    but dotproduct is not cross product ... so i think your attempts are misguided

  17. Loser66
    • one year ago
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    The goal of cross product is to find out a vector perpendicular to the given vectors, right?

  18. Loser66
    • one year ago
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    Hence if we can't find it out by cross while dot can help us, why not?

  19. amistre64
    • one year ago
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    that is an application of it yes, but i am not sure if that is the only reason for it.

  20. Loser66
    • one year ago
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    Do you know something about flat ?

  21. Loser66
    • one year ago
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    I read a paper but I don't get what it is.

  22. amistre64
    • one year ago
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    'flat' doest ring any bells ...

  23. Loser66
    • one year ago
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    ok, thanks for the help. :)

  24. amistre64
    • one year ago
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    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|

  25. amistre64
    • one year ago
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    there are going to be an infinite number of vectors in R^4 that are perpendicular to a R^2 object.

  26. Loser66
    • one year ago
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    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

  27. amistre64
    • one year ago
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    |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 ...

  28. amistre64
    • one year ago
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    find "a" vector that mets the condition it amounts to a system of n equations, in more than n unknowns.

  29. amistre64
    • one year ago
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    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

  30. amistre64
    • one year ago
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    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

  31. amistre64
    • one year ago
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    or (a,b,0,0,0) or .... whatever is simplest for you to deduce

  32. amistre64
    • one year ago
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    just a thought, but it may be related to finding pivot points for those nxm matrix setup to evaluate them

  33. amistre64
    • one year ago
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    -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 :)

  34. Loser66
    • one year ago
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    I will. Thank you so much. :)

  35. Loser66
    • one year ago
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    |dw:1444576604879:dw|

  36. Loser66
    • one year ago
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    |dw:1444580267634:dw|

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