(Another) Hard Vectors Question (2 Viewers)

[mreditor16]

LOL please help. Don't where to start tbh.

 

[awesome-0_4000]

10 vectors with components all 0 except for one which is 5, so 5e1, 5e2, etc where ei are the standard basis vectors in R^10. You can then see that there is no 11th vector that satisfies the given conditions.

 

[mreditor16]

10 vectors with components all 0 except for one which is 5, so 5e1, 5e2, etc where ei are the standard basis vectors in R^10. You can then see that there is no 11th vector that satisfies the given conditions.

Sorry I should have clarified. I got the first part out, and I am stuck on the 11th vector part. According to the answers, there is a 11th vector :/

 

[VBN2470]

Hope it helps

 

[photastic]

Question OP, why doesn't your tutor help you during tutorials?

Not sure if my solution is valid but VBN clarifies things better.

Vector a = (β,β,...β) ε R^10
Find β such that
|ai - a| = 5sqrt2
ai - a = (-β,..-β,α-β,-β,-β)
|ai - a|= 9β^2 +(α-β)^2 = 50
9β^2 +(5-β)^2 = 50
Then solve for β or whatever you called the parameter.

 

[VBN2470]

Question OP, why doesn't your tutor help you during tutorials?

Not sure if my solution is valid but VBN clarifies things better.

Vector a = (β,β,...β) ε R^10
Find β such that
|ai - a| = 5sqrt2
ai - a = (-β,..-β,α-β,-β,-β)
|ai - a|= 9β^2 +(α-β)^2 = 50
9β^2 +(5-β)^2 = 50
Then solve for β or whatever you called the parameter.

Your working seems correct, but I think OP would also want a justification as to why each component of the 11th vector need to be equal, otherwise you wouldn't really be getting the same distance for EVERY pair. To make things simpler, try the exact same case for R^2 and R^3 (where you would look for a 3rd and 4th vector respectively) and the same method would apply for R^10.

 

[mreditor16]

Question OP, why doesn't your tutor help you during tutorials?

Not sure if my solution is valid but VBN clarifies things better.

Vector a = (β,β,...β) ε R^10
Find β such that
|ai - a| = 5sqrt2
ai - a = (-β,..-β,α-β,-β,-β)
|ai - a|= 9β^2 +(α-β)^2 = 50
9β^2 +(5-β)^2 = 50
Then solve for β or whatever you called the parameter.

My tutor's 1 minute explanation didn't make sense at all. Hence, me asking it on BOS.

 

[mreditor16]

but thanks VBN

 

[VBN2470]

My tutor's 1 minute explanation didn't make sense at all. Hence, me asking it on BOS.

Do you remember what your tutor said?

 

[mreditor16]

Do you remember what your tutor said?

Tbh not much, because I think we only had a few minutes left and they miraculously seemed to pluck the quadratic out of nowhere, and how they got said quadratic was really confusing.

 

[VBN2470]

Tbh not much, because I think we only had a few minutes left and they miraculously seemed to pluck the quadratic out of nowhere, and how they got said quadratic was really confusing.

Fair enough then, I asked because then maybe I could have reasoned out what they meant and tried to deduce some easier way of solving this, especially if it took them around a minute to do (unless they knew the q beforehand, and pulled the quadratic of out nowhere). Wonder if anyone else can shed some light on an easier method for this question? That would be good.

 

[InteGrand]

Another method is just substitute in the ansatz (where there are 10 a's), and solve for a and hope you get a real solution (which you do in the end), rather than proving you need all components same.

 

[VBN2470]

Another method is just substitute in the ansatz (where there are 10 a's), and solve for a and hope you get a real solution (which you do in the end), rather than proving you need all components same.

But how would you know all components need to be the same? Why can't they be different?

 

[InteGrand]

I didn't say they can't be different. You could just seek one that has all same components and see if you can solve the resulting equation.

 

[VBN2470]

I didn't say they can't be different. You could just seek one that has all same components and see if you can solve the resulting equation.

Fair enough

 

[mreditor16]

Anyone have a 'nicer' solution? Also, keep in mind guys that this is a first year, first semester question. I don't believe I am familiar with "ansatz" for example.

 

[InteGrand]

Ansatz just means a guess (http://en.wikipedia.org/wiki/Ansatz).

 

[VBN2470]

I guess a more intuitive method would be to consider the fact the each of the 10 basis vectors (in R^10) are all orthogonal (perpendicular) to each other and each lie on its corresponding 'axis', so the only way in which an 11th such vector can be equidistant to these vectors is if it is inclined 45 degrees to each vector i.e. some scalar multiple of v = (1, 1, ... ,1)^T, so we would only need to find the scaling factor such that the distance is preserved.