Parallelogram Vector Proof: Using Vector Addition and Triangle Relationships

[duldin]

Homework Statement

Incl. diagrams:
http://i.imgur.com/w6Exj.png

Homework Equations

Just vector addition (?)

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) + 0.5aFrom 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 :)

 

[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)

 

[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 :(

 

[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?

 

[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 separate sides. But, what does the a and b are not parallel clue tell you? Please help!

 

[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

 

[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!

 

[duldin]

Triangles EDQ and EBA appear to be similar triangles.

Does this imply,

BE/ED = AE/EQ

?

 

[jing2178]

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

 

[jing2178]

yes it does

 

[rupink06]

It means that m and n are equal...

 

[jing2178]

Triangles EDQ and EBA appear to be similar triangles.

Does this imply,

BE/ED = AE/EQ

?

yes it does.

 

[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.

 

[duldin]

yes it does.

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

 

[jing2178]

Sorry I had to go and do some gardening

You have

Rearranging I get:
xa + 0.5ya - 0.5a = yb - xb

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.

 

[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 ?

:(

 

[jing2178]

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

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.

 

[duldin]

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

Thanks heaps.

 

[jing2178]

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