# Parallelogram Vector Proof

1. Mar 31, 2012

### duldin

1. The problem statement, all variables and given/known data
Incl. diagrams:
http://i.imgur.com/w6Exj.png

2. Relevant equations

3. The attempt at a solution
a.) ED = x(BD)
ED = x(a+b)

b) EQ = y(AQ)
= y(b-0.5a)

From the smaller triangle:
ED = EQ + QD
ED = [y(b-0.5a)] + (0.5a)
I am not sure if this is what part b is asking, but it's all I can figure?

c) From the last two parts,
ED = ED
x(BD) = y(b-0.5a) + (0.5a)
x(a+b) = y(b-0.5a) + 0.5a

From the 'hint' given, I think I'm meant to somehow remove y, to then solve and acquire x=1/3. I cannot work out how though.

Thanks :)

Last edited: Mar 31, 2012
2. Mar 31, 2012

### jing2178

part (a) asks you to express ED in terms of x and a and b. You have yet to do this. When you have then follow (b) and (c)

3. Mar 31, 2012

### duldin

Ah sorry, I have done that with pen I just forgot to type those lines out.

ED = x(BD)
ED = x(a+b)

Substitute in and I end up with this:
ED = ED
x(a+b) = y(b-0.5a) + 0.5a

I can't work out how to get rid of the y to show x=1/3 :(

4. Apr 1, 2012

### jing2178

Put all the as on one side of the equation and all the bs on the other. Given the hint that a and b are not parallel what does this tell you?

What can you say about triangles EDQ and EBA? What does this tell you about x and y?

5. Apr 1, 2012

### rupink06

Jing, could you please explain further? I'm doing the same question and still can't see how to get rid of Y. I got the As and Bs to seperate sides. But, what does the a and b are not parallel clue tell you? Please help!

6. Apr 1, 2012

### jing2178

With m and n scalars and a and b vectors such that a and b are not parallel and ma=nb how is this possible?

Remember a=pb p not zero implies a is parallel to b

7. Apr 1, 2012

### rupink06

I still don't get it... I understand that a and b are not parallel if ma=mb, but how do I do this mathematically.

I currently have this equation:

a(2x+y-1)=b(2y-x)

I get that (2x+y-1) doesn't equal (2y-x), but how do I integrate the non-parallel factor into this equation???

I really hate this question and its due tomorrow! D:

More help please! Really appreciate it!

8. Apr 1, 2012

### duldin

Triangles EDQ and EBA appear to be similar triangles.

Does this imply,

BE/ED = AE/EQ

?

9. Apr 1, 2012

### jing2178

Think about (m-n)*5 equals 0 what can you say about m and n?

10. Apr 1, 2012

### jing2178

yes it does

11. Apr 1, 2012

### rupink06

It means that m and n are equal...

12. Apr 1, 2012

### jing2178

yes it does.

13. Apr 1, 2012

### rupink06

Jing, could you please, just explain the answer, so I can work backwards to found out how its done myself? Running tight on time here.

14. Apr 1, 2012

### duldin

Thanks. Not sure if that goes where I want it too but I tried this one first:
BA/QD = BE/ED

Substituting each of those vectors in terms of a and b yields the desired x=1/3... without using y! It's magic. Since it doesn't use y I'm not sure if it constitutes a correct response to the question considering the lack of y... but it works!

Would you mind clarifying where you are going with the parallel lines bit. Rearranging I get:
xa + 0.5ya - 0.5a = yb - xb
I see it doesn't make sense since no matter what y and x are, they're still just scalars so that equation is stating they are parallel. Were you just suggesting that was the completely wrong way to do it? :P

15. Apr 1, 2012

### jing2178

Sorry I had to go and do some gardening

You have

so (x + 0.5y -0.5)a = (y-x)b

but if ma=nb and a and b are not parallel then m=n=0 is only possibility.

I was hoping that if you saw x must equal y from the similar triangles you would get the (y-x) part would equal 0 and be able to proceed from there.

16. Apr 1, 2012

### duldin

Thanks for the help.

I see where you are coming from, but I don't see how you boil that down to x=1/3 ???

:(

17. Apr 1, 2012

### jing2178

It follows that x+0.5y-0.5=0 ........(1) and y-x=0 ............(2)

so from (2) y=x

substitute into (1)

x+0.5x-0.5=0 mult by 2

2x + x -1 =0 you can finish from here.

Last edited: Apr 1, 2012
18. Apr 1, 2012

### duldin

Ah, dead easy. Simultaneous equations in a vectors question... good stuff :thumbsup:

Thanks heaps.

19. Apr 1, 2012

### jing2178

To progress in maths you will need to understand that the various topics you have learnt and will learn do not exist in isolation. Anything you have learnt can be of use in any future question.