Use cross-product to find vector in R^4 that is orthogonal

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  • #1
DryRun
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Homework Statement
http://s2.ipicture.ru/uploads/20111115/ltM3iwGZ.jpg

The attempt at a solution

Please correct me if i'm wrong in my assumptions:
R^4 means that i need to find a vector that exists in 4 dimensions, meaning 4 rows.

I am trying desperately to visualise this problem, with 4 axis since it involves 4 dimensions, but i can't. How would 4 dimensions appear? I learned about 3 dimensions, using the right-handed system and the three axis; x, y, z, each axis perpendicular to each other.

My basic understanding of this problem is that i have to solve this vector equation:
axbxc, where a is the first column vector, b, is the second, and so on.
However, in my undergraduate syllabus, it says it's called dyadics and it's beyond the scope of my programme.
So, i'm thinking (axb)xc but this is not the same thing! unless i'm mistaken?
Also, if the vector i need to find, is perpendicular to any two from the given vectors, it should naturally be perpendicular to the 3rd given vector, right?
 
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  • #2
D H
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Where did you get this problem? If it is from a class you are taking, the text or lectures should have supplied the information needed to solve the problem. If you just happened across it, you don't have the necessary context.

For now, I'll assume it came from a class. How did the text/lectures define the cross product in RN? (Hint: It involves N-1 vectors.)
 
  • #3
DryRun
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It's based on class lectures and the problem is from the tutorial. However, my notes don't contain anything about R^4 or higher dimensions. The maximum described is up to R^3. Maybe i need to read about it, as the lecturer always tells us to go refer in the recommended books; Calculus: One and Several Variables by Salas and Hille and also, Linear Algebra by Strang.
I checked in the Calculus book (which the class notes very closely follows) but it doesn't seem to have anything related to 4 dimensions.
The lectures defined the cross product in R^3 as axb, quite simply.
Then, axb = c, where c is the vector normal to the plane containing vectors a and b.

There is also this part about identities:
ax(bxc)=(a.c)b-(a.b)c
(axb)xc=(c.a)b-(c.b)a
(axb).(cxd)=(a.c)(b.d)-(a.d)(b.c)
 
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  • #4
D H
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One way to express the cross product of two vectors a and c in R3 is in the form of a determinant,

[tex]\mathbf a \times \mathbf b =
\begin{vmatrix}
\hat x & \hat y & \hat z \\
a_x & a_y & a_z \\
b_x & b_y & b_z \end{vmatrix}
[/tex]

Generalizing this to N dimensions would require N-1 vectors to create an analogous NxN "matrix" (it isn't really a matrix, hence the scare quotes):

[tex]\mathbf a_1 \times \mathbf a_2 \times \cdots \times \mathbf a_{N-1} =
\begin{vmatrix}
\hat e_1 & \hat e_2 & \cdots & \hat e_n \\
a_{1,1} & a_{1,2} & \cdots & a_{1,N} \\
a_{2,1} & a_{2,2} & \cdots & a_{2,N} \\
\vdots & \vdots & & \vdots \\
a_{N-1,1} & a_{N-1,2} & \cdots & a_{N-1,N} \end{vmatrix}
[/tex]

The vector generated by this determinant will be zero if the N-1 vectors are not linearly independent, non-zero and orthogonal to each of the N-dimensional cross product is between N-1 vectors.

This extension of the determinant form of the cross product in R3 to RN is one way to generalize the cross product.
 
  • #5
DryRun
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Thanks for your help, D H. I understand now. It's actually quite simple, as it seems to be an extension of the same method used for cross-product in R^3.
 
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  • #6
DryRun
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Please bear with me, but can i use this formula instead?

[URL]http://upload.wikimedia.org/wikipedia/en/math/0/9/4/0946149d6b5f3a8f6a4c54593492a388.png[/URL]

Or maybe i can use this as well?

[URL]http://upload.wikimedia.org/wikipedia/en/math/3/c/2/3c264630c3ca28f0c0b627027d106789.png[/URL]

If i can, then how do i know how to group the vectors?
 
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  • #7
D H
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The vector triple product for R3 is not the same as the determinant-based extension of the cross product to R4.

What is the motivation for this hmmmm ... stuff? Has your instructor said what they are trying to do, and why? It is pretty non-standard, which is why you are not getting much help.
 
  • #8
DryRun
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The syllabus in my undergraduate class is based on Linear Algebra and Calculus. Just a few weeks ago, we started Calculus in a single variable, and currently we just finished Limits and continuity. I received the tutorial from the lecturer in my email, as the maths test is scheduled for next week, so i'm trying to work through it. I just follow the tutorials mainly, as the notes are just main points with reference (heavily) to the recommended books. Really, the entire question (it's actually the last one in the tutorial) is exactly as in the screen-shot shown in the first post of this thread.

To go back to the problem, here is how i worked it out, although i'm not sure if it's correct.

Determinant of:
w x y z
1 2 3 4
2 1 3 1
-1 0 1 -1

I can solve this 4x4 determinant. The w, x, y, z are vectors with a hat to indicate unit vector. So, i would assume that's the answer?
 
  • #9
D H
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Correct.

You can double check (you should double check!) that the computed determinant is in fact normal to each of the provided vectors. That will be the case if you computed the determinant correctly. If you get a non-zero inner product between the computed determinant and any of the provided vectors it means you made a mistake in either the computation of the determinant or in the computation of the inner product.
 
  • #10
Ray Vickson
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There is a theorem stating that if you want to have a "cross-product" f(u,v) that, for any two vectors u and v in R^n, returns another vector in R^n, and has the usual algebraic properties, then n must be 3 or 7. So there are no cross products in dimensions 2, 4, 5, 6, and 8 or more.

RGV
 

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