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    (Original post by master y)
    Find the direction vector of AB by doing b-a, 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?
    I may have done it wrong but that doesn't get 3:2 :/
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    (Original post by adam17)
    I may have done it wrong but that doesn't get 3:2 :/
    It does, basically you get root(54):root(24). These are in the ratios of 3 to 2.

    Spoiler:
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    AB = (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.
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    (Original post by master y)
    It does, basically you get root(54):root(24). These are in the ratios of 3 to 2.

    Spoiler:
    Show
    AB = (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.
    Oh right thanks, I think I managed to work out BC wrong when I tried it.....

    Thank you for your help
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    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.
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    (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.
    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 x-coord 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
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    (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 x-coord 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
    ahh ok. Thanks very much
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    I feel really worried about core 3, feel really unprepared
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    Can anybody help me out on Q4ii of the June 08? http://www.ocr.org.uk/Images/62967-q...hematics-4.pdf

    Don't really understand the mark scheme!
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    (Original post by erniiee)
    Can anybody help me out on Q4ii of the June 08? http://www.ocr.org.uk/Images/62967-q...hematics-4.pdf

    Don't really understand the mark scheme!
    Probably one of the hardest C4 questions i've seen, I much prefer the dot product/solving for intersections compared to this stuff.

    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 (3-2t)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!
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    (Original post by erniiee)
    Can anybody help me out on Q4ii of the June 08? http://www.ocr.org.uk/Images/62967-q...hematics-4.pdf

    Don't really understand the mark scheme!

    (Original post by Genesis2703)
    X
    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.
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    (Original post by Oxy-moron Prodigy)
    X.
    Tbh I think I dislike it more due to the fact that it was a weird vectors question that doesn't come up much. And it is rare a question quite early completely stumps me when I originally do it. I sort of get the method now. It does seem a bit strange though, because (wait my bad diagram now)

    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."
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    Nice to find this thread! How are you guys preparing
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    Can someone help me on 8 (iv) ?
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    (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 (3-2t)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!
    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

    (Original post by Oxy-moron 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.
    Could you explain what's in bold - the position vector of P is the equation you constructed?

    (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."
    This is the part I'm struggling to understand!

    (Original post by strikerextreme)
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    Can someone help me on 8 (iv) ?
    Oh this question bugged me big time when I tried it, quote me if you manage to find a good explanation please!
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    (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!
    You need to remember what r is, r is an equation in terms of a position vector and a direction vector, there is a parameter/constant on the direction vector. This means if lets say you put t = 1 into your equation of r, you would get a POINT on that line. So what we are saying is that P is a point on that line, but we do not know what value of t is needed in order to get that point. Hence we have to express P in terms of the whole of equation r with the t still in as a variable, (but of course you actually carry out the addition, you don't leave P as the 2 vectors in part one you merge them. e.g. is r = (1,2,3) + t(1,1,1) the point P is at (1+t, 2+t, 3+t). Then since O is a (0,0,0) the direction OP is (1+t, 2+t, 3+t).

    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!
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    Does anyone have any sites with practice questions? Thank you!
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    (Original post by JoshL123)
    Does anyone have any sites with practice questions? Thank you!
    Do you mean past papers? or have you already done them all haha! If you mean random non-paper questions then I can't help
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    (Original post by Genesis2703)
    Do you mean past papers? or have you already done them all haha! If you mean random non-paper questions then I can't help
    Yeah I have :L. I have been going pretty keen on this as I need to get an A* to meet my offer, and maths is my only hope! I'm doing the papers again, but just like random questions. By the way, as you seem to know everything :P, just wondering what your opinions were on integrating stuff like Sin(ax)Cos(bx)? I saw it in a textbook and it literally stumped me :/
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    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...
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    (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!
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