What is the Angle Between Vectors Method?

AI Thread Summary
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.
t_n_p
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Homework Statement



A little left of field this question..

http://img455.imageshack.us/img455/4531/bantuzfy5.jpg

The Attempt at a Solution



I'm unsure more with the wording of this question if anything rather than the method of how to go about it. How would I go about finding the vector that represents a "half Inuit, half Bantu" population?
 
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Take the average of the two columns.
 
Seems pretty logical/right, but best to wait for another person to confirm this is 100% right.

Thanks christianjb!
 
Personally I would go for the geometric mean of the 2 columns, rather than arithmetic mean (assuming the numbers represent probabilities?), but It could still be correct to say arithmetic mean.
 
Since they are talking about these as vectors, I would interpret "half Inuit, half Bantu" to mean (1/2)\vec{I}+ (1/2)\vec{B}, adding 1/2 of each vector. of course, that is the same as taking the arithmetic average (mean).
 
I guess it's the average of two then. Thanks guys
 
Managed to figure that one out after using the mean of the two but got stuck soon after on this question.

Amongst all possible combinations that are mix of Inuit and Bantu, find the mix
that is closest to the English population. (Hint: Set up things such that the
infinitely many different possible mixed populations correspond to a vector that depends on a variable, say t.)

With the hint, I'm thinking I should use Gaussian elimination somehow? Is there a better method?
 
Halls, I can't see your Latex graphic!
 
I would interpret a "mix" of Inuit and Bantu vectors as t\vec{I}+ (1-t)\vec{B} where t is a number between 0 and 1. That will give the "infinitely many different possible mixed populations" they are talking about. Find the t that minimizes the distance between that and \vec{E}.
 
  • #10
hmmm, not sure how I would go about finding a t value that minimizes distance...
 
  • #11
bump*********
 
  • #12
Do you know how to find the distance between two vectors: ||u- v||.
That will be quadratic in t and then complete the square.
 
  • #13
HallsofIvy said:
Do you know how to find the distance between two vectors: ||u- v||.
That will be quadratic in t and then complete the square.

I don't really understand, can you elaborate?
 
  • #14
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  • #15
woah!
So first I find the distance between the two vectors, but which two vectors in particular? My common sense tells me between vector E and vector (tI - (1-t)B. But how do I interpret (tI - (1-t)B?

Slightly confused! :confused:
 
  • #16
I'm not sure calculating the magnitude of the difference between the vectors will give the right solution... it is the angle that needs to be minimized...

calculate:
H = tI + (1-t)B

t is just a scalar, write out I and B in (x1,x2,x3,x4) form... then you should be able to calculate H and write it in (x1,x2,x3,x4) form.

Then do the dot product between E (english) and H... you have a formula for dot product that relates it to the magnitudes and the angle between the vectors...
 
  • #17
t_n_p said:
woah!
So first I find the distance between the two vectors, but which two vectors in particular? My common sense tells me between vector E and vector (tI - (1-t)B.
Correct.
But how do I interpret (tI - (1-t)B?

Slightly confused! :confused:
Weighted average of I and B (weight = t). A.k.a. convex linear combination of I and B.
 
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  • #18
learningphysics said:
I'm not sure calculating the magnitude of the difference between the vectors will give the right solution... it is the angle that needs to be minimized...

calculate:
H = tI + (1-t)B

t is just a scalar, write out I and B in (x1,x2,x3,x4) form... then you should be able to calculate H and write it in (x1,x2,x3,x4) form.

Then do the dot product between E (english) and H... you have a formula for dot product that relates it to the magnitudes and the angle between the vectors...
I am thinking... "If there is any justice in the world" then the two should give the same solution. Why would they be different?
 
  • #19
EnumaElish said:
I am thinking... "If there is any justice in the world" then the two should give the same solution. Why would they be different?

Because the vectors don't have the same magnitude...

If we're using 2d vectors for example... you can construct two vectors at a fixed angle, and arbitrarily change the length of one of the vectors keeping the other fixed... the third side (representing the difference in the two vectors) will increase in magnitude, but the angle is fixed...

If you restrict the two sides being the same length then it doesn't matter... but in this case the magnitudes are different.
 
  • #20
Okay, I am convinced. Why do you think angle is the right approach, not distance? Why should magnitude not matter?
 
  • #21
EnumaElish said:
Okay, I am convinced. Why do you think angle is the right approach, not distance? Why should magnitude not matter?

The intial question mentions the genetic distance \theta as the angle between the two vectors... So I was just going with that...
 
  • #22
learningphysics said:
I'm not sure calculating the magnitude of the difference between the vectors will give the right solution... it is the angle that needs to be minimized...

calculate:
H = tI + (1-t)B

t is just a scalar, write out I and B in (x1,x2,x3,x4) form... then you should be able to calculate H and write it in (x1,x2,x3,x4) form.

Then do the dot product between E (english) and H... you have a formula for dot product that relates it to the magnitudes and the angle between the vectors...

Ok I've just got a small question.
After plugging in I and B in the form (x1,x2,x3,x4), I get this result..

http://img67.imageshack.us/img67/5144/untitledgg9.jpg

What do I do next? I can't just expand out because those are commas, not +, - etc...:confused:
 
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  • #23
t_n_p said:
Ok I've just got a small question.
After plugging in I and B in the form (x1,x2,x3,x4), I get this result..

http://img67.imageshack.us/img67/5144/untitledgg9.jpg

What do I do next? I can't just expand out because those are commas, not +, - etc...:confused:

t and 1-t are just scalars... it's just multiplication of a vector by a scalar... eg 2(1,1,1,1) = (2,2,2,2)

Hope that you've learned about vectors and dot product etc... because this problem involves that.
 
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  • #24
Yeah, just had a small mind block!
would the second section become (0.1-0.1t, 0.08-0.08t, etc) ?
 
  • #25
t_n_p said:
Yeah, just had a small mind block!
would the second section become (0.1-0.1t, 0.08-0.08t, etc) ?

Yup, that's right.
 
  • #26
Ok, now I've done that.
H = (0.29t, 0t, 0.03t, 0.67t) + (0.1-0.1t, 0.08-0.08t, 0.12-0.12t, 0.69-0.69t)

How do I dot product E with H now that H is in this (slightly weird) form?
 
  • #27
t_n_p said:
Ok, now I've done that.
H = (0.29t, 0t, 0.03t, 0.67t) + (0.1-0.1t, 0.08-0.08t, 0.12-0.12t, 0.69-0.69t)

How do I dot product E with H now that H is in this (slightly weird) form?

Add up those two parts of H...
 
  • #28
ah, so H = (0.29t+0.1-0.1t, 0t+0.08-0.08t, etc...)
 
  • #29
Yup. Now try to use the dot product with E now... it's going to get a little tricky... you also have a formula that relates dot products with angles... that's what you need here.

The idea is to minimize the angle with E.
 
  • #30
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?
 
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  • #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.....
 

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