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    (Original post by Genesis2703)
    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!

    Thank you so much! I understand it now - P has to satisfy the equation so you just need to find the value of the parameter. Thanks for explaining it in detail bro!

    I'll have a look at my god awful working for that question 8, I did very strange things
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    Wonder if someone could explain question 5ii

    http://papers.xtremepapers.com/OCR/M...tion_Paper.pdf

    Thanks
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    Can someone explain vectors to me. I don't understand at all
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    How is everyone learning their trig identities. Really struggling to remember them. Does anyone have a method ?? X
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    (Original post by SexyAndIKnowIt.)
    How is everyone learning their trig identities. Really struggling to remember them. Does anyone have a method ?? X
    Well learn the simple cos2x + sin2x = 1
    then know you have to divide through by cos2x to give you the identity
    1 +tan2x =sec2x
    also know you have to divide the original identity by sin2x to give you the identity
    cot2x + 1 = cosec2x

    For the double angle ones, using them is the best way to remember but if you really can't then I would suggest just working them out from scratch like cos(x+x)=cosxcosx - sinxsinx
    then manipulate them as you wish
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    (Original post by JoshL123)
    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 :/
    There are two ways of doing such integrals if they aren't in an obvious form, for example sin2xcos2x you could recognise as 0.5sin4x due to double angle formulae but if it was sin2xcos3x then it is not as simple.

    if the integral is called I, if you do integrating by parts twice you should end up with I = (some number) -kI where k is a constant (positive or negative). In other words after two iterations of parts the thing you have to integrate (in the 2nd part of the parts formula) will be identical to thing thing you wanted to integrate in the 1st place, so you can rearrange and solve the integral as you would an equation.

    The 2nd way is well, using your addition/subtraction angle formulae. sin(A+B) = sinAcosB + sinBcosA and sin(A-B) = sinAcosB - sinBcosA

    So sin(A+B) + sin(A-B) = 2sinAcosB, if your integral was sin3xcos2x, with this new identity you will see that sinAcosB = 0.5(sin(A+B) + sin(A-B)) and in this integral you want A to be 3x and B to be 2x so sin3xcos2x = 0.5(sin(3x+2x) + sin(3x-2x)) = 0.5(sin(5x) + sin(x))

    to which in this form you can integrate normally!

    These integrals are very rare in C4 papers, examiners would be very mean to put them in the C4 paper without a part prior showing you the identity needed, I doubt they'd make you think of the identity yourself.
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    Wonder if someone could explain question 5ii

    http://papers.xtremepapers.com/OCR/M...tion_Paper.pdf

    Thanks
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    Whats the raw mark usually needed for 90ums in c4 and c3???
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    (Original post by max25)
    Whats the raw mark usually needed for 90ums in c4 and c3???
    about 64ish
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    (Original post by adam17)
    Wonder if someone could explain question 5ii

    http://papers.xtremepapers.com/OCR/M...tion_Paper.pdf

    Thanks
    after you work out the expansion just replace x with (x+x^3) and expand/simplify.
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    how do you work out a unit vector? what is it?
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    (Original post by master y)
    how do you work out a unit vector? what is it?
    Is this the question in that paper.... I know which question I think you are on about haha.

    A unit vector is a vector for which it's magnitude is 1, i.e. |a| = 1, I assume you know that you do the square root of I^2 + J^2 + K^2, and yeah that should equal 1
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    Hi I was wondering if anyone had the mark scheme to jan 2013?? Would be very much appreciated!!
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    (Original post by megsy14)
    Hi I was wondering if anyone had the mark scheme to jan 2013?? Would be very much appreciated!!
    Do you have the question paper? I've been looking for it for ages!
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    (Original post by adam17)
    Do you have the question paper? I've been looking for it for ages!
    I only have a paper copy sorry!
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    (Original post by Genesis2703)
    Is this the question in that paper.... I know which question I think you are on about haha.

    A unit vector is a vector for which it's magnitude is 1, i.e. |a| = 1, I assume you know that you do the square root of I^2 + J^2 + K^2, and yeah that should equal 1
    could you help me with this question? The points A and B have position vectors (4,-2,3) and (1,1,-3). Find a unit vector parallel to the line AB....

    actually, worked it out! Thanks anyway Genesis
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    (Original post by master y)
    could you help me with this question? The points A and B have position vectors (4,-2,3) and (1,1,-3). Find a unit vector parallel to the line AB....

    actually, worked it out! Thanks anyway Genesis
    Hi, sorry but what did you get as your answer?
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    (Original post by megsy14)
    Hi, sorry but what did you get as your answer?
    1/root 54 (-3,3,-6) ... you?
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    (Original post by master y)
    1/root 54 (-3,3,-6) ... you?
    Yep same, thanks!
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    Any one has the core 4 Jan 2013 question paper please?
 
 
 
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