Maths Paper II Q c ii (Vectors) 1999

Splicer

Hi,
don't know how to do Maths Paper II Q 2 c ii 1999 on vectors.
Anyone?
THANKS!:)

mathew

its not too difficult, just tedious.. Use the dot product rule.

For the first part I assume you let r = ai+bj and s=xi+yj
and work from there

for the next part it is just unit perpendicular vectors.

I would write out the whole answer but I dont have the time.

Your teacher might be able to help!!!!!

ashyle

Splicer wrote:

Hi,
don't know how to do Maths Paper II Q 2 c ii 1999 on vectors.
Anyone?
THANKS!:)

those jaysus vectors. become a linear programming person like me

mp3guy

www.examinations.ie

go to examination materials archive, then marking schemes

:rolleyes:

cocoa

marking schemes only go back as far as 01, he wants 99...

Splicer

cheers guys
i looked at the solutions in the celtic press books but i didnt get them!... hmm, worrying!
i ll talk to my teacher. should help

Tomlowe

theres one vectors question in the past somewhere that's absolutely mad, my teacher refuses to do it in class, to quote "Whoever set that question should be strung up with piano wire"

i think its that one

mp3guy

cocoa wrote:

marking schemes only go back as far as 01, he wants 99...

I stand corrected. Is there anywhere to get those old ones? A site with 10 years+ of past marking schemes would generate so much revenue from advertising each year.

MetalHead

I learned how to do the Linear Programming Question by myself in about 3 days because the questions that come up are identical to the previous years, maybe that's something to consider. I've got probably the Worst Maths teacher in Galway, she's teaching us Vectors so I just decided to move to Linear Programming!

ZorbaTehZ

Anyone still looking for the answers to this question?

E92

My Maths teacher told us that that part of the Q was only 5 out of 50 marks. I'll type up the answer if anyone wants it.

cocoa

There's actually two ways of answering it, the dot product formula and pure logic. IMO the pure logic is much easier if you can grasp it.

If you divide a vector by it's modulus then you have a unit vector.
The same is true of the perpendicular (only you have the perpendicular unit vector)
If you add or subtract two vectors of equal length (unit vectors, obviously qualify) then the resultant will make the same angle with both vectors.
If you were to turn both original vectors 90degrees then the similarity of angle will remain.
The length of ad, defined by t is irrelevant as we are concerned with angle.

Oh and the whole question looks a lot neater (with either method) if you redefine a as the origin, so ab becomes b and ac becomes c and ad becomes d.

ZorbaTehZ

The pure logic one isn't accepted as an answer.
Redefining a as the origin isn't accepted either because that's a change of angle.

The easiest way is putting the angles equal using the dot product rule.

ZorbaTehZ

mp3guy wrote:

I stand corrected. Is there anywhere to get those old ones? A site with 10 years+ of past marking schemes would generate so much revenue from advertising each year.

QFT

dan719

ZorbaTehZ wrote:

The pure logic one isn't accepted as an answer.
Redefining a as the origin isn't accepted either because that's a change of angle.

The easiest way is putting the angles equal using the dot product rule.

I reckon the pure logic would be accepted, as long as it was stated as above, but you cannot change the angle by redefining origin. Of course there would probably be a hefty row at the marking conference.:D

Stupid typos!lol

ZorbaTehZ

In any and all the vector questions, logic arguments and drawings are strictly a no-no - it has to be shown using the formulas.

EDIT:Typo

dan719

I have this row with my teacher all the time. And have often handed up 'pure logic' answers and been given full marks.
Also to show that most people resort to 'pure logic' as a matter of course.
What would be your first method of proving the following (k)(k+1) is an even number where k is greater then or equal to one. There are about ten different ones but what would you use first.

ZorbaTehZ

dan719 wrote:

I have this row with my teacher all the time. And have often handed up 'pure logic' answers and been given full marks.
Also to show that most people resort to 'pure logic' as a matter of course.
What would be your first method of proving the following (k)(k+1) is an even number where k is greater then or equal to one. There are about ten different ones but what would you use first.

Mate, reread my post:

ZorbaTehZ wrote:

In any and all the vector questions, logic arguments and drawings are strictly a no-no - it has to be shown using the formulas.

I'm talking about vector questions. I never said anything about the other questions on the paper.

dan719

ZorbaTehZ wrote:

Mate, reread my post:



I'm talking about vector questions. I never said anything about the other questions on the paper.

And I am saying that I have handed up pure logic answers in vectors.

ZorbaTehZ

Lucky you then, because (s)he isn't supposed to accept them.

dan719

ZorbaTehZ wrote:

Lucky you then, because (s)he isn't supposed to accept them.

And you know this how? And if it makes you feel better vectors is my extra question anyway.

ZorbaTehZ

That's makes me feel much better mate, thanks.
Because someone who has been correcting Maths exam papers for the last 3 years told me.

dan719

Oh I guess you must be right so, because my maths teacher has only been teaching for thirty years and correcting for more then twenty of them. Who would you believe?:D

ZorbaTehZ

My teacher, thanks.

EDIT: Typo

cocoa

Aside from the leaving cert, it is a true answer. In regards to the leaving cert, you might lose marks but you could quite possibly get them back at the recheck.

Why can't you redefine a as the origin? You aren't moving it, you're just changing its name...

ZorbaTehZ

You are changing the angles.
Consider a triangle with x, y, z angles - if you were to move one of the vertices and keep the other two vertices stationary, you are going to change all the angles. Just draw a picture and you can see it quite clearly.

ron-burgandy

ZorbaTehZ wrote:

In any and all the vector questions, logic arguments and drawings are strictly a no-no - it has to be shown using the formulas.

EDIT:Typo

2005 Paper 2 question 2(A)

ZorbaTehZ

When I said that, I meant when it says show/prove as is the context of the original question, that this thread was made about. That question you refer to is finding a point.
Guess I need to be more specific in my posts, else I'll be responding to these tedious/petty examples, ad nauseum.

cocoa

ZorbaTehZ wrote:

You are changing the angles.
Consider a triangle with x, y, z angles - if you were to move one of the vertices and keep the other two vertices stationary, you are going to change all the angles. Just draw a picture and you can see it quite clearly.

ok, I'm going to say this again, slowly.

you don't need to move a
you just need to redefine it.

ZorbaTehZ

What you originally said: (in case you've forgotten)

cocoa wrote:

Oh and the whole question looks a lot neater (with either method) if you redefine a as the origin, so ab becomes b and ac becomes c and ad becomes d.

Redefining as the origin is the only way to simplify it. You can't do that.
As for redefining to something else, that's pointless since it doesn't change the question hardly at all.
Comprende?

Tomlowe

redefining is the same as moving, but it doesnt matter in this particular question because there is no reference to the origin: all the vectors are transformations from one to the other and all the vertices of angles are at a, so redefining a as the origin doesn't change any angles.

P(1) says: 1(2) is even, which is true
assume k(k+1) is even
(k+1)(k+2)
=2(k+1) + k(k+1)

2(k+1) is even as it is divisible by 2
we assumed that k(k+1) is even

the sum of two even numbers is an even number, so 2(k+1) + k(k+1) is even.

Since the truth of P(k) implies the truth of P(k+1) and the statement is true for n=1, the statement is true for all n E N (less 0)

As far as I can see if you make a logical progression without using numbers to prove a statement in general then apply it to the specific in question, that should be absolutely fine.

cocoa

ok, I'm going to spell this right out to you because you're clearly not catching on easily.

Imagine a, b, c and d are all points.
Now move a to o
Now move b, c and d under the same transformation.

The only thing that's changed is the question looks a ****load neater. All of the lines mentioned in the question originate at a.

sniped by tomlowe, *agrees*

Tomlowe

also im pretty sure that the marking schemes are copyrighted, so you cant just stick them up on the internet and make a profit, its somewhere on the examinations website

edit: "Users may retrieve examination material solely for their own personal, non-commercial use, as specified at (i) above, and may download the material to their own hard disc or send it to a printer solely for that purpose. They may not otherwise copy, modify, or distribute the examination material, or publish, broadcast, transmit, or otherwise distribute any portion of this material without the express written authorisation of the State Examinations Commission. Any unauthorised use is strictly prohibited. The State Examinations Commission permit no unauthorised modifications, adaptations or translations of the examination material."

dan719

Tomlowe wrote:

redefining is the same as moving, but it doesnt matter in this particular question because there is no reference to the origin: all the vectors are transformations from one to the other and all the vertices of angles are at a, so redefining a as the origin doesn't change any angles.

P(1) says: 1(2) is even, which is true
assume k(k+1) is even
(k+1)(k+2)
=2(k+1) + k(k+1)

2(k+1) is even as it is divisible by 2
we assumed that k(k+1) is even

the sum of two even numbers is an even number, so 2(k+1) + k(k+1) is even.

Since the truth of P(k) implies the truth of P(k+1) and the statement is true for n=1, the statement is true for all n E N (less 0)

As far as I can see if you make a logical progression without using numbers to prove a statement in general then apply it to the specific in question, that should be absolutely fine.

No you misunderstood me. The question itself is piss easy, and it can quite easily be done using induction but why not use so called pure logic.

Statement one. Any even number multiplied by an odd number is always = to even number. e.g 6x3=18

if k is even, k+1 is odd but from above odd by even = even

if k is odd , k+1 is even and again using above. This is just as acceptable as and in my view more elegant then using induction. In this case induction just seems too heavy handed as it were.:D

ZorbaTehZ

cocoa wrote:

ok, I'm going to spell this right out to you because you're clearly not catching on easily.

Imagine a, b, c and d are all points.
Now move a to o
Now move b, c and d under the same transformation.

The only thing that's changed is the question looks a ****load neater. All of the lines mentioned in the question originate at a.

sniped by tomlowe, *agrees*

You realize that you said redefine a as the origin, not use transformations?

Tomlowe

yep, sure, but for leaving cert purposes id always use induction (and would always be told to anyway)

Pure logic should be fine, but seems like you're making a hell of a lot of trouble for yourself

Tomlowe

zorbatehz: same thing

cocoa

ZorbaTehZ wrote:

You realize that you said redefine a as the origin, not use transformations?

you realise that it's quite annoying that while everyone else I talk to understands the word "redefine" to mean imagine everything moves with it/imagine it was always there but for you I have to go to the trouble of explaining it slowly?

ZorbaTehZ

Letting a be the origin.

As opposed to:

Letting a be the origin and moving all the other points through the same transformation.

How is that the same thing?

Tomlowe

redefine a as the origin means move the origin to a, not move a to the origin