Finding Orthogonal Vectors in R^5 without Cross Product

phil ess
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Homework Statement



Given vectors a=(1,-1,0,2,1) b=(3,1,-2,-1,0) and c=(1,5,2,4,-4), which are mutually orthogonal, find a system of linear equations that a vector x must satisfy so it is orthogonal to a, b, and c.

Homework Equations



None I think.

The Attempt at a Solution



This is part A of the problem, after which I just have to solve the system and I'm confident I can do the rest once I have it, but I don't know where to start. I tried using the equations:

x1-x2+3x4+x5=0
3x1+x2-2x3-x5=0
x1+5x2+2x3+4x4-4x5=0

And then just solving the matrix, which gives me
x=(0,0,0,0,0)+s(-8,2,-11,5,0)+t(-5,5,-6,0,2)

But I'm pretty sure this doesn't answer the problem. I don't know how to create a system that gives orthogonal solutions, and the above solution isn't even a vector as far as I know. How do I start this problem? Any tips are much appreciated! Thanks.
 
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You are doing exactly the right thing, except there seems to be some typos in your equations. How did you get the 3 in the first one and why is there an x5 and not an x4 in the second one. Your result isn't 'a' vector, it's a two parameter family of vectors, which is exactly what you should get, but the numbers aren't quite right either.
 
I understand the problem now, and I went back and reworked everything carefully and ended up with the right solution. It was actually much easier than I thought at first. Thanks a lot!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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