What is the Angle Between Vectors Method?

In summary, the conversation is discussing a question about finding the vector that represents a "half Inuit, half Bantu" population. After some debate, it is determined that the best approach is to take the average of the two columns, which is the same as taking the arithmetic mean. Then, a new question arises about finding the mix of Inuit and Bantu populations that is closest to the English population. The hint given is to set up a vector that depends on a variable, t, and find the t value that minimizes the distance between that vector and the English vector. It is suggested to find the distance using the dot product and using the formula that relates it to magnitudes and angles between vectors. The conversation ends with a
  • #36
Man, finding the magnitude of H is going to be really messy...
 
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  • #37
t_n_p said:
Man, finding the magnitude of H is going to be really messy...

Yeah... it might simplify to something clean... :-p
 
  • #38
hmmm, cleaning that up leads to...

http://img406.imageshack.us/img406/2203/untitledod2.jpg

which leads to..

http://img406.imageshack.us/img406/3793/untitledgh4.jpg

Now what?! :confused:
 
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  • #39
t_n_p said:
hmmm, cleaning that up leads to...

http://img406.imageshack.us/img406/2203/untitledod2.jpg

which leads to..

http://img406.imageshack.us/img406/3793/untitledgh4.jpg

Now what?! :confused:

Ah... I just thought of something... you know that cos0 = 1... 0 is the smallest angle (no need for derivatives.) So what you're doing is finding the t that ends up giving an angle of 0.

so you can set the left side = 1 and just solve for t... still a little tricky, but you'll end up with a quadratic formula so it should be ok...
 
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  • #40
Ok, so I bring the denominator over to one sides, square both sides and then bring everything over to one side to give...

http://img374.imageshack.us/img374/1821/untitledxk3.jpg

Now I solve using quadratic formula?
 
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  • #41
t_n_p said:
Ok, so I bring the denominator over to both sides, square both sides and then bring everything over to one side to give...

http://img374.imageshack.us/img374/1821/untitledxk3.jpg

Now I solve using quadratic formula?

Yup. :smile:
 
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  • #42
t = 10.935 or 0.2823
I suppose I should take the lesser value and substitute it into
http://img67.imageshack.us/img67/5144/untitledgg9.jpg
 
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  • #43
t_n_p said:
t = 10.935 or 0.2823
I suppose I should take the lesser value and substitute it into
http://img67.imageshack.us/img67/5144/untitledgg9.jpg

exactly. t=0.2923 is the one you want since you're looking for a number between 0 and 1.
 
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  • #44
Done! The mix that is closest to the english population is..

http://img514.imageshack.us/img514/5941/untitledhu6.jpg

Now I need to give a mathematical proof that the angle between two 4-dimensional vectors v
and w is well-defined for all possible choices of the two vectors.
 
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  • #45
t_n_p said:
Done! The mix that is closest to the english population is..

http://img514.imageshack.us/img514/5941/untitledhu6.jpg

Now I need to give a mathematical proof that the angle between two 4-dimensional vectors v
and w is well-defined for all possible choices of the two vectors.

Cool!

Do you have any ideas for the proof?
 
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  • #46
learningphysics said:
Cool!

Do you have any ideas for the proof?

Not particularly..:bugeye:
 
  • #47
Look at the formula you posted for the angle:

http://img385.imageshack.us/img385/308/untitledjq2.jpg

That is how the angle is defined...

You need to show that no matter what 4 vectors you have for v or w, this formula will always give you an angle...

You might have learned theorems in class, or they might be in your textbook to assist with this...
 
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  • #48
Well, my textbook isn't particularly helpful. Tutor also said this is quite a hard problem.
 
  • #49
t_n_p said:
Well, my textbook isn't particularly helpful. Tutor also said this is quite a hard problem.

It is a hard problem. But the first step isn't that hard:

Suppose I gave this formula:

[tex]\theta = cos^{-1} x[/tex]

For what values of x is [tex]\theta[/tex] well-defined?
 
  • #50
-1 to 1.....
 
  • #51
t_n_p said:
-1 to 1.....

Exactly.

So using that same idea, when is [tex]\theta[/tex] well defined in the dot product formula?
 
  • #52
-1 to 1...LOL
 
  • #53
t_n_p said:
-1 to 1...LOL

But what has to be between -1 and 1, for [tex]\theta[/tex] to be defined? :wink:

It was x in that formula I gave... but in the dot product formula it is...?
 
  • #54
for theta to be defined, http://img356.imageshack.us/img356/5728/untitledjq2kn7.jpg has to be between -1 and 1
 
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  • #55
t_n_p said:
for theta to be defined, http://img356.imageshack.us/img356/5728/untitledjq2kn7.jpg has to be between -1 and 1

Exactly. :smile: So that's what you need to prove, to show that [tex]\theta[/tex] is well defined... you need to show that that expression is between -1 and 1 for any two 4-vectors v and w.

It's a tough problem if you're solving from scratch... There a theorem called the "Cauchy-Schwarz inequality" that shows that this is true almost immediately... Have you covered it in your course?
 
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  • #56
Havn't learned the theorem. If it's essenetial to the question, I don't know why they ask non relevant questions...:confused:
 
  • #57
t_n_p said:
Havn't learned the theorem. If it's essenetial to the question, I don't know why they ask non relevant questions...:confused:

Hmmm... it might also be called by another name...

Basically the Cauchy-Schwarz inequality says

[tex]|\overrightarrow{v}.\overrightarrow{w}| \le |\overrightarrow{v}||\overrightarrow{w}|[/tex]

and from that the result you want follows immediately...

you've probably got something like that explained or proven in your text. Look it up in your text or online...

did your tutor say anything else about this problem?
 
  • #58
I'll catch you later t_n_p. I got to :zzz:
 
  • #59
Never seen anything remotely like that before! My tutor just said try best, don't worry too much if I can't do it.
 
  • #60
learningphysics said:
I'll catch you later t_n_p. I got to :zzz:

no worries, thanks alot!
 

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