What is the Angle Between Vectors Method?

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The discussion centers on finding the vector that represents a "half Inuit, half Bantu" population, with participants debating whether to use the arithmetic or geometric mean. The consensus is that the arithmetic mean, represented as (1/2)I + (1/2)B, is appropriate for combining the vectors. The main challenge is to find a mix of these populations that is closest to the English population vector, leading to the formulation of a variable t that represents the mix. Participants discuss using the dot product to minimize the angle between the resulting vector and the English vector, ultimately leading to the conclusion that the angle should be minimized rather than the distance. The conversation concludes with a focus on ensuring that the angle between the vectors is well-defined within the context of their mathematical properties.
  • #31
t_n_p said:
Ok, after cleaning up the dot product I get..

E·H = 0.0146t + 0.4874

Now I apply the formula

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

But how exactly do I minimize the angle?

cool. I'd rewrite as:
cos\theta=\frac{\overrightarrow{v}.\overrightarrow{w}}{|\overrightarrow{v}||\overrightarrow{w}|}

What do you usually do to get maximums or minimums?
 
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  • #32
find derivative and equate to zero and proof max/min?
 
  • #33
t_n_p said:
find derivative and equate to zero and proof max/min?

yup. that's the idea.
 
  • #34
So derive E·H = 0.0146t + 0.4874?

well d/dt (E·H) = 0.0146

To test for max/min I use double derivative test.
which leads to 0 (point of inflexion)

?
 
  • #35
t_n_p said:
So derive E·H = 0.0146t + 0.4874?

well d/dt (E·H) = 0.0146

To test for max/min I use double derivative test.
which leads to 0 (point of inflexion)

?

No, you want \frac{d\theta}{dt} to be 0 because \theta is what you're minimizing.

So you need to use that dot product formula, with E instead of v, and H instead of w... simplify that first before worrying about the derivative... you've already got the numerator... the denominator will be a little messy to calculate...
 
  • #36
Man, finding the magnitude of H is going to be really messy...
 
  • #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:

\theta = cos^{-1} x

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

Exactly.

So using that same idea, when is \theta 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 \theta 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 \theta 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

|\overrightarrow{v}.\overrightarrow{w}| \le |\overrightarrow{v}||\overrightarrow{w}|

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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