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    Jan 13 paper and MS
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  1. File Type: pdf 4724-01Jan13 (1).pdf (201.5 KB, 198 views)
  2. File Type: pdf 4724_MS_Jan13.pdf (162.0 KB, 217 views)
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    (Original post by Genesis2703)
    ...
    Hey man, could you help me on part i of question 9 on the June 08? http://www.ocr.org.uk/Images/62967-q...hematics-4.pdf
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    (Original post by erniiee)
    Hey man, could you help me on part i of question 9 on the June 08? http://www.ocr.org.uk/Images/62967-q...hematics-4.pdf

    B is at an intersection with the x-axis, meaning that y = 0, A it at the highest points possible (highest y-value possible). Does that help?
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    (Original post by Genesis2703)
    B is at an intersection with the x-axis, meaning that y = 0, A it at the highest points possible (highest y-value possible). Does that help?
    Well I understand that, and I also appreciate what the mark scheme says in terms of the answer, but as we've been given the parametric equations, do we literally just have to assume the graph has the same "limits" in terms of range and domain as the parametric equations?

    The way I found the values was assuming the graph followed the same pattern as the sine graph in terms of highest value and value at y=0, but only because that's the trig function that was in the parametric equations..so lets say those were cos instead of sine but everything else the same (except for the graph shifting to the right appropriately) we'd assume theta = 0 and 3pi over 2?
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    (Original post by Genesis2703)
    B is at an intersection with the x-axis, meaning that y = 0, A it at the highest points possible (highest y-value possible). Does that help?
    I saw your post on the FP2 thread, congrats on slaying the beast

    hows ur c4 revision going?
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    hi guys...you know when a question which involves expanding a binomial series....usually the question right after is...for what x values is this valid for?...can someone please explain the approach to answering that sort of question. thanks.
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    (Original post by mrmccarl)
    hi guys...you know when a question which involves expanding a binomial series....usually the question right after is...for what x values is this valid for?...can someone please explain the approach to answering that sort of question. thanks.
    http://examsolutions.net/maths-revis...tutorial-1.php
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    can someone help me with this vectors question please? (june 2012 core 4 not mei Q10)

    lines L1 and L2 have vector equations:

    r= -i+2j+7k + t(2i+2j+k) and r=2i+9j-4k + s(i+3j-2k)
    respectively. The point A has coordinates (-3,0,6) relative to the origin O

    i)) show that A lies on L1 and that OA is perpendicular to L1
    ii) show that the line through O and A intersects L2
    iii) Given that the point of intersection in part (ii) is B find the ratio (magnitude of OA): (magnitude of BA)
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    (Original post by bellaanne)
    can someone help me with this vectors question please? (june 2012 core 4 not mei Q10)

    lines L1 and L2 have vector equations:

    r= -i+2j+7k + t(2i+2j+k) and r=2i+9j-4k + s(i+3j-2k)
    respectively. the point A lies on L1 and has coordinates (-3,0,6) relative to the origin O

    i)) show that A lies on L1 and that OA is perpendicular to L1
    ii) show that the line through O and A intersects L2
    iii) Given that the point of intersection in part (ii) is B find the ratio (magnitude of OA): (magnitude of BA)
    Which part?
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    (Original post by joostan)
    Which part?

    Part ii)
    Im unsure what to do. Do we have to find the vector equation of A?

    I know it's position vector is the coordinates that are given: (-3,0,6)
    but Im unsure what it's direction vector is. I was thinking of using the same direction vector of the line L2 for A but im not 100% sure

    Thanks
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    (Original post by bellaanne)
    Part ii)
    Im unsure what to do. Do we have to find the vector equation of A?

    I know it's position vector is the coordinates that are given: (-3,0,6)
    but Im unsure what it's direction vector is. I was thinking of using the same direction vector of the line L2 for A but im not 100% sure

    Thanks
    The direction vector of OA is
    \begin{pmatrix} -3 \\ 0 \\ 6 \end{pmatrix}
    Does this help?
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    (Original post by joostan)
    The direction vector of OA is
    \begin{pmatrix} -3 \\ 0 \\ 6 \end{pmatrix}
    Does this help?
    what? i thought that was it's position vector
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    (Original post by bellaanne)
    what? i thought that was it's position vector
    Yes, but it is also the direction vector of A from the origin if that makes sense?
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    Ohhhh ok, i get you now. I just wanted to quickly ask that if a point (position vector) LIES on a line does that mean that it has the SAME direction vector as the line?
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    (Original post by bellaanne)
    Ohhhh ok, i get you now. I just wanted to quickly ask that if a point (position vector) LIES on a line does that mean that it has the SAME direction vector as the line?
    Not necessarily, but if the line passes through the origin then yes.
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    Could someone explain 4ii on June 2008 paper please? I hate vectors!
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    (Original post by m00c0w)
    Could someone explain 4ii on June 2008 paper please? I hate vectors!
    the position vector P will have co-ordinates (x,y,z)

    therefore OP = p - o ------> so P remains unchanged (x,y,z)

    if P lies on the line then you can say:

    (x,y,z) = (3, 2, 3) + t(-2,1,1)

    therefore you know the values of x, y and z

    you are then told that they are perpendicular so the scalar prod = 0

    using that information you can deduce that -2x+y+z = 0

    you already know what x,y and z are so just sub in and solve to find t; then you can work out the co-ordinates of position vector P.
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    (Original post by a10)
    the position vector P will have co-ordinates (x,y,z)

    therefore OP = p - o ------> so P remains unchanged (x,y,z)

    if P lies on the line then you can say:

    (x,y,z) = (3, 2, 3) + t(-2,1,1)

    therefore you know the values of x, y and z

    you are then told that they are perpendicular so the scalar prod = 0

    using that information you can deduce that -2x+y+z = 0

    you already know what x,y and z are so just sub in and solve to find t; then you can work out the co-ordinates of position vector P.
    Surely you would need to work out t to be able to find x, y and z. How would you do that?
    Thanks!
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    (Original post by m00c0w)
    Surely you would need to work out t to be able to find x, y and z. How would you do that?
    Thanks!
    well you know what x, y and z equal to, so sub those values in and you can work out t then use that value of t to work out the co-ordinates of the point P.
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    (Original post by a10)
    well you know what x, y and z equal to, so sub those values in and you can work out t then use that value of t to work out the co-ordinates of the point P.
    Wait, how do we know what x,y and z are equal to? Other than:
    x=3-2t
    y=2+t
    z=3+t
 
 
 
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