dejan91's questions

[dejan91]

OK well the question:
The sides of triangle ABC are represented by the vectors A>B = a, B>C = b, and C>A = c, such that a + b = -c (sorry i have no idea how to type in vectors, so A>B is AB with arrow an at the top like this →).

Prove the cosine rule, |c|² = |a|² + |b|² - 2|a||b|cos B.

Any help would be much appreciated.

[Mao]

let a and b be two position vectors from O. Let , thus forming a triangle.









QED

[BlueYoHo]

oooooooh yeah now i see. wow... what a coinidence. I'd just gotten stuck on this question and decided to take a break/look here.

so a.a for example is the same as |a|^2 ?

[Mao]

[monokekie]

haha, i am stuck on this questions too..

thanks mao lol

[dejan91]

Thanks for help man. However, I asked my teacher today, and this is what he said to do:

c.c = |c|²
=a.a + a.b + a.b + b.b
=a.a + b.b + 2a.b
=a.a + b.b + 2|a||b|cos B

Now, find the positive supplementary angle.
|c|² = a.a + b.b + 2|a||b|cos(180 - B)

Since 180 - B is in second quadrant, cos(180 - B) will be negative. Therefore,
|c|² = a.a + b.b - 2|a||b|cosB
       = |a|² + |b|² - 2|a||b|cosB.

Of course, correct me if I'm wrong 

[Mao]

ahhh, silly me. you are right. I completely forgot to take into account the direction of a and b.

However, changing angles can get messy. It is easier to define a and b to be pointing outwards from, hence c=a-b. I have edited my proof above.

[dejan91]

Thanks Mao fo help anyway (Y)

Also, this isn't on vector proofs, but cbf starting new thread. Had this question that I just left ages ago till now..

If a = 2i - 3j + k and b = -2i + 3j + k,
find a unit vector perpendicular to both a and b.

[/0]

Let the vector perpendicular to both and be

Then, by the dot product:





Adding the two equations, we find that and so .

So a possible vector could be

And the unit vector would be

[dejan91]

/0 thanks, however your answer is slightly off the books answer. I don't have the book with me now, but I will post it up later.

Also, this isn't a vector proof, and may not even be in the specialist maths course, but I am interested in knowing wether this actually works or not (note this isn't my work, and I had no idea where else to post this):

Prove that if x = 0.9999...
       x = 1.


10x = 9.9999...

10x - x = 9x = 9

Therefore, x = 1,
and x = 0.9999...

Is this correct??

[/0]

I don't know how valid that proof is, but I prefer using geometric series:

0.999... = 0.9 + 0.09 + 0.009 + ...

a = 0.9, r = 0.1

[TrueTears]

/0 thanks, however your answer is slightly off the books answer. I don't have the book with me now, but I will post it up later.

Also, this isn't a vector proof, and may not even be in the specialist maths course, but I am interested in knowing wether this actually works or not (note this isn't my work, and I had no idea where else to post this):

Prove that if x = 0.9999...
       x = 1.


10x = 9.9999...

10x - x = 9x = 9

Therefore, x = 1,
and x = 0.9999...

Is this correct??

yes that proof is correct, and it is perfectly valid.

[kamil9876]

/0 thanks, however your answer is slightly off the books answer. I don't have the book with me now, but I will post it up later.

Also, this isn't a vector proof, and may not even be in the specialist maths course, but I am interested in knowing wether this actually works or not (note this isn't my work, and I had no idea where else to post this):

Prove that if x = 0.9999...
       x = 1.


10x = 9.9999...

10x - x = 9x = 9

Therefore, x = 1,
and x = 0.9999...

Is this correct??

yes that proof is correct, and it is perfectly valid.

The geometric series method is "more" valid since it defined the number 0.9999... as the limit of a sum. This 'proof' merely treats the number as an infinite string of 9's after the decimal rather than treating the number in terms of a quantity. Using such syntax based reasoning can lead incorrect results:

p=2*2*2*2...
2p=2(2*2*2*2...)
2p=p
2p-p=0
hence p=0
and 2*2*2*2*2...=0

However sometimes a correct result is yielded:



Let





and so x=2

So 2 is the answer. Both proofs used the same method; perform an operation on both sides and use the infinite repitition of the process to get the value of x or p in some some solvable equation. However one is wrong, one is right and limits fill this void, just like the limit based approach of the geometric series does. However the proof involving 0.999... can easily be made more formal and it is in fact analogous to the derivation of the sum of a geometric series (ask for the details if u want, i feel my post is already long enough LOL)

[Mao]

[kamil9876]

Wouldn't the proof of this premise also use geometric series? Hence 0.333..*3=0.999 would be synonymous to 3*((3/10)+(3/100)+(3/1000)...)=(9/10)+(9/100)+(9/100)...

And so we would have to some how work out the sum on the right, which is the limit of the sum of a geometric series, or use some other method which would probably just be the same thing in disguise.