# Angle between vectors method

1. Aug 2, 2007

### t_n_p

1. The problem statement, all variables and given/known data

A little left of field this question..

http://img455.imageshack.us/img455/4531/bantuzfy5.jpg [Broken]

3. 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?

Last edited by a moderator: May 3, 2017
2. Aug 2, 2007

### christianjb

Take the average of the two columns.

3. Aug 2, 2007

### t_n_p

Seems pretty logical/right, but best to wait for another person to confirm this is 100% right.

Thanks christianjb!

4. Aug 2, 2007

### Gib Z

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.

5. Aug 2, 2007

### HallsofIvy

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).

6. Aug 3, 2007

### t_n_p

I guess it's the average of two then. Thanks guys

7. Aug 6, 2007

### t_n_p

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?

8. Aug 7, 2007

### t_n_p

Halls, I can't see your Latex graphic!

9. Aug 8, 2007

### HallsofIvy

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. Aug 8, 2007

### t_n_p

hmmm, not sure how I would go about finding a t value that minimizes distance....

11. Aug 10, 2007

### t_n_p

bump*********

12. Aug 10, 2007

### HallsofIvy

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. Aug 10, 2007

### t_n_p

I don't really understand, can you elaborate?

14. Aug 10, 2007

### EnumaElish

Last edited: Aug 10, 2007
15. Aug 10, 2007

### t_n_p

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!

16. Aug 10, 2007

### learningphysics

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. Aug 11, 2007

### EnumaElish

Correct.
Weighted average of I and B (weight = t). A.k.a. convex linear combination of I and B.

Last edited: Aug 11, 2007
18. Aug 11, 2007

### EnumaElish

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. Aug 11, 2007

### learningphysics

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. Aug 11, 2007

### EnumaElish

Okay, I am convinced. Why do you think angle is the right approach, not distance? Why should magnitude not matter?