I may have done it wrong but that doesn't get 3:2 :/(Original post by master y)
Find the direction vector of AB by doing ba, and then find the magnitude of this (add the squares of x,y,z then square root). Do the same for BC.
does this work?
You are Here:
Home
> Forums
>< Study Help
>< Maths, science and technology academic help
>< Maths
>< Maths Exams

OCR C4 (not mei) 18th June 2013 revision watch
Announcements

 Follow
 41
 07062013 17:39

master y
 Follow
 2 followers
 0 badges
 Send a private message to master y
 Thread Starter
Offline0ReputationRep: Follow
 42
 07062013 17:47
(Original post by adam17)
I may have done it wrong but that doesn't get 3:2 :/
Spoiler:ShowAB = (3, 3, 6) BC=(2,2,4)
you then do root(9+9+36) and same for BC, and these are the magnitudes of the two vectors. 
 Follow
 43
 07062013 18:36
(Original post by master y)
It does, basically you get root(54):root(24). These are in the ratios of 3 to 2.
Spoiler:ShowAB = (3, 3, 6) BC=(2,2,4)
you then do root(9+9+36) and same for BC, and these are the magnitudes of the two vectors.
Thank you for your help 
 Follow
 44
 08062013 21:25
I have a question about C3 but I can't find a proper thread for it so I thought I'd post it here in hope that someone can help
On one of the questions in the Jan 13 paper it asks you to find the range so I differentiated it but it turned out I was supposed to complete the square :/ . When finding the range of some of the more complicated f(x) questions how do you know whether to differentiate or complete the square?
This might be a really stupid question but I'm really struggling with/hating these sorts of questions atm. 
 Follow
 45
 09062013 15:04
(Original post by kitkat95)
I have a question about C3 but I can't find a proper thread for it so I thought I'd post it here in hope that someone can help
On one of the questions in the Jan 13 paper it asks you to find the range so I differentiated it but it turned out I was supposed to complete the square :/ . When finding the range of some of the more complicated f(x) questions how do you know whether to differentiate or complete the square?
This might be a really stupid question but I'm really struggling with/hating these sorts of questions atm.
1) complete the square and the number outside the bracket is your minimum value for y.
2)differentiate and find The xcoord of the stationary point, then sub that value of x into the original equation to find your Lowest value of y.
both methods give you the same answer 
 Follow
 46
 09062013 21:16
(Original post by dadams2)
Yeah, there are two ways to do the question:
1) complete the square and the number outside the bracket is your minimum value for y.
2)differentiate and find The xcoord of the stationary point, then sub that value of x into the original equation to find your Lowest value of y.
both methods give you the same answer 
tinkerbell_xxx
 Follow
 40 followers
 15 badges
 Send a private message to tinkerbell_xxx
Offline15ReputationRep: Follow
 47
 10062013 13:01
I feel really worried about core 3, feel really unprepared

 Follow
 48
 10062013 13:20
Can anybody help me out on Q4ii of the June 08? http://www.ocr.org.uk/Images/62967q...hematics4.pdf
Don't really understand the mark scheme! 
Genesis2703
 Follow
 11 followers
 3 badges
 Send a private message to Genesis2703
Offline3ReputationRep: Follow
 49
 10062013 14:30
(Original post by erniiee)
Can anybody help me out on Q4ii of the June 08? http://www.ocr.org.uk/Images/62967q...hematics4.pdf
Don't really understand the mark scheme!
Since the question mentions perpendicular you know you are going to use the dotproduct = to 0 i.e. a.b = 0 (since cos(90) = 0 on the RHS).
But what is a and b you ask? Remember they are directions vectors, so you need two directions. The 1st direction is the direction AB (i.e. the part of your answer to part i that is the direction vector, it has the constant t (or s etc.) in front of it).
You need a second direction though, you know P is on AB, and you want OP to be perpendicular. since O is at (0,0,0) when working out the direction vector OP, it is the same as the position vector of P. As OP = (position P  position O and O = 0).
What is P. well you are trying to find that so you need a general form. You know P is on AB correct, so that means the position of P can be represented as your WHOLE answer of i) and what we need to do is find the value of t (in front of the direction vector) to give us what P actually is.
Sorry if I worded that badly, but jn short the direction vector of AB is 2i + j + k OR 2i  j k (both are correct, but use the one you put in part i) of course). and the direction of OP is your whole of i) i.e. the WHOLE line of your answer to i) which has multiple possibilities again. One of which is (32t)i + (2+t)j + (3+t)k
You then use the dot product with these two vectors, find what t is to make it = 0, and then sub that value of t into part i) to see what position P is in!
Sorry I rambled on a bit. I hope that made sense though! 
Oxymoron Prodigy
 Follow
 2 followers
 2 badges
 Send a private message to Oxymoron Prodigy
Offline2ReputationRep: Follow
 50
 10062013 14:44
(Original post by erniiee)
Can anybody help me out on Q4ii of the June 08? http://www.ocr.org.uk/Images/62967q...hematics4.pdf
Don't really understand the mark scheme!
(Original post by Genesis2703)
X
From part i) I got the equation for AB as (3i + 2j + 3k) + s(2i + 1j + 1k)
As OP is said to be perpendicular to AB, we know the dot product of the two lines has to = 0. a.b. = 0
(This might not be technically correct but I think of it this way) OP does not have a separate direction vector as it is relative to the origin and it's position vector is simply ( (3  2s)i + (2 + 1s)j + (3 + 1s)k ) which is the equation for AB 'combined' if you like. It's much easier to see if you write the equations out as column vectors (downwards).
a.b. = 0 so we multiply out the direction vectors to get 2(3  2s) + (2 + s) + (3 + s) = 0.
Expanding out the brackets gives 6 + 4s + 2 + s + 3 + s = 0.
6s = 1, therefore s = 1/6
You have found s and you had previously found the position vector of P so simply subsitute the value of s into the equation.
This gives (8/3i + 13/6j + (19/6k)
I've checked the mark scheme and this is the correct answer. 
Genesis2703
 Follow
 11 followers
 3 badges
 Send a private message to Genesis2703
Offline3ReputationRep: Follow
 51
 10062013 15:15
(Original post by Oxymoron Prodigy)
X.
if AB was like and we left OP be
>
^


etc. we are finding the angle between the arrow head of OP but the tail end of AB (the head is intersecting the tail I mean). While if you look at all the diagrams in the C4 textbook you find the angle between the two tails
e.g.
^

>
so yeah, why do you look at OP and not PO, to get a similar shape to the 2nd diagram which is a horrid drawing of the textbook version haha.
Oh and the examiner report agrees it says:
"Very few managed any coherent attempt at this part. A vectorial effort was required to indicate that OP and AB were perpendicular but few realised that the direction vector of OP was the whole of the RHS of their equation in part (i) and the direction vector of AB was the portion after ‘t’ parameter in its equation; the scalar product of these two, equated to 0, soon gave t = 1/6 or 1/6 or 5/6 or 5/6[depending on which version of r = (a or b) + t(b – a or −ba ) was given in part (i)] and this, substituted into the part (i) equation, produced the required position vector."Last edited by Genesis2703; 10062013 at 15:17. 
 Follow
 52
 10062013 22:59
Nice to find this thread! How are you guys preparing

strikerextreme
 Follow
 1 follower
 0 badges
 Send a private message to strikerextreme
Offline0ReputationRep: Follow
 53
 11062013 02:00
Can someone help me on 8 (iv) ? 
 Follow
 54
 11062013 14:29
(Original post by Genesis2703)
You need a second direction though, you know P is on AB, and you want OP to be perpendicular. since O is at (0,0,0) when working out the direction vector OP, it is the same as the position vector of P. As OP = (position P  position O and O = 0).
Just to clarify, OP is also a line, so has an equation with a position vector and direction vector like AB?
What is P. well you are trying to find that so you need a general form. You know P is on AB correct, so that means the position of P can be represented as your WHOLE answer of i) and what we need to do is find the value of t (in front of the direction vector) to give us what P actually is.
Sorry if I worded that badly, but jn short the direction vector of AB is 2i + j + k OR 2i  j k (both are correct, but use the one you put in part i) of course). and the direction of OP is your whole of i) i.e. the WHOLE line of your answer to i) which has multiple possibilities again. One of which is (32t)i + (2+t)j + (3+t)k
You then use the dot product with these two vectors, find what t is to make it = 0, and then sub that value of t into part i) to see what position P is in!
Sorry I rambled on a bit. I hope that made sense though!
Also, sorry for the stupid extra question in bold, but it is a line with an equation right?! This question is driving me crazy haha
(Original post by Oxymoron Prodigy)
Dam Genesis beat me to it. I don't think it is one of the hardest C4 questions, once you understand it it's not too difficult.
From part i) I got the equation for AB as (3i + 2j + 3k) + s(2i + 1j + 1k)
As OP is said to be perpendicular to AB, we know the dot product of the two lines has to = 0. a.b. = 0
(This might not be technically correct but I think of it this way) OP does not have a separate direction vector as it is relative to the origin and it's position vector is simply ( (3  2s)i + (2 + 1s)j + (3 + 1s)k ) which is the equation for AB 'combined' if you like. It's much easier to see if you write the equations out as column vectors (downwards).
a.b. = 0 so we multiply out the direction vectors to get 2(3  2s) + (2 + s) + (3 + s) = 0.
Expanding out the brackets gives 6 + 4s + 2 + s + 3 + s = 0.
6s = 1, therefore s = 1/6
You have found s and you had previously found the position vector of P so simply subsitute the value of s into the equation.
This gives (8/3i + 13/6j + (19/6k)
I've checked the mark scheme and this is the correct answer.
(Original post by Genesis2703)
Oh and the examiner report agrees it says:
"Very few managed any coherent attempt at this part. A vectorial effort was required to indicate that OP and AB were perpendicular but few realised that the direction vector of OP was the whole of the RHS of their equation in part (i) and the direction vector of AB was the portion after ‘t’ parameter in its equation; the scalar product of these two, equated to 0, soon gave t = 1/6 or 1/6 or 5/6 or 5/6[depending on which version of r = (a or b) + t(b – a or −ba ) was given in part (i)] and this, substituted into the part (i) equation, produced the required position vector."

Genesis2703
 Follow
 11 followers
 3 badges
 Send a private message to Genesis2703
Offline3ReputationRep: Follow
 55
 11062013 20:36
(Original post by erniiee)
Thanks for the reply. I'm a bit confused about what I've put in bold  why is the position of P represented by the whole of what r equals? Is it simply because its a point on the line?
Also, sorry for the stupid extra question in bold, but it is a line with an equation right?! This question is driving me crazy haha
This is the part I'm struggling to understand!
Oh this question bugged me big time when I tried it, quote me if you manage to find a good explanation please!
And yeah the examiners report just says what I have just said, P is a point on the line, so it can be represented algebraically as a point on the line with the t as a variable, then you have to find t and voila!
On the 2nd question 8iv. Well the last part is basically just the 1st 3 parts over again just with some actual values, you know that for points P Q of the form specified in the introduction, the normal to the curve at P is p, the chord joining the points is ii) and the general equation for it is in part iii).
So now in iv) we have a situation where P and Q are in essence R and S, BUT the normal at S meets T as well in the same way, so you have to do it twice. we know (8,8) must be in the form of P and Q specified at the start for this situation to exist. so with P being (2p^2, 4p) we can easily see that 8 = 4p so p = 2, now if we treat points S and q (looking at equation iii)) we can sub p into it to find q, 2^2 + 2q + 2 = 0, q = 3. BUT S is related to T in the same way, so we have to put that 3 back into iii) again to find the value q2 for T, which is (3)^2 3q + 2 so q2 = 11/3
Then you put q2 into (2t^2, 4t) and get T. The most confusing part of the question is the change in letters, basically P is R, Q is S, and T is a 3rd point related to S in the same way R is related to S! 
 Follow
 56
 11062013 22:49
Does anyone have any sites with practice questions? Thank you!

Genesis2703
 Follow
 11 followers
 3 badges
 Send a private message to Genesis2703
Offline3ReputationRep: Follow
 57
 11062013 22:50
(Original post by JoshL123)
Does anyone have any sites with practice questions? Thank you! 
 Follow
 58
 11062013 22:54
(Original post by Genesis2703)
Do you mean past papers? or have you already done them all haha! If you mean random nonpaper questions then I can't help 
 Follow
 59
 12062013 00:54
2 papers done today just gotta mark them, btw that question 8iv posted by someone above also threw me I had no idea where to start these parametric questions can get real tricky...

strikerextreme
 Follow
 1 follower
 0 badges
 Send a private message to strikerextreme
Offline0ReputationRep: Follow
 60
 12062013 01:42
(Original post by erniiee)
Thanks for the reply. I'm a bit confused about what I've put in bold  why is the position of P represented by the whole of what r equals? Is it simply because its a point on the line?
Also, sorry for the stupid extra question in bold, but it is a line with an equation right?! This question is driving me crazy haha
Could you explain what's in bold  the position vector of P is the equation you constructed?
This is the part I'm struggling to understand!
Oh this question bugged me big time when I tried it, quote me if you manage to find a good explanation please!
Reply
Submit reply
Related discussions:
 OCR A2 (NonMEI) C4 Exam  18th June 2014
 OCR MEI FP2 (18th June) Exam Thread
 OCR (non mei) M2 Wednesday 18th May 2016
 OCR F324June 2013 Chemistry
 Edexcel C3,C4 June 2013 Thread
 OCR Physics A G485  Frontiers of Physics  18th June 2015
 A2 Biology OCR June 2015 Revision Thread
 AQA Physics PHYA5  Thursday 18th June 2015 [Exam ...
 Ocr mei statistics 1 exam thread (24/05/13)
 AQA A2 Mathematics MPC3 Core 3  Friday 5th June 2015 ...
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
 SherlockHolmes
 Notnek
 charco
 Mr M
 TSR Moderator
 Nirgilis
 usycool1
 Changing Skies
 James A
 rayquaza17
 RDKGames
 randdom
 davros
 Gingerbread101
 Kvothe the Arcane
 The Financier
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 Reality Check
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 LeCroissant
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Moltenmo
 Labrador99
Updated: May 7, 2014
Share this discussion:
Tweet