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Maths C4 Help!

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    Can anyone explain how to get the coordinates of D?

    The answer says that Lambda is 3, so therefore the coordinates for D is (8, -19, 11)

    I don't understand how they arrived at lambda being 3...

    Thanks in advance!
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    DOES ANYONE DO MEI M2?
    Please pm me.
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    (Original post by albus)
    DOES ANYONE DO MEI M2?
    Please pm me.
    :confused: :rolleyes: Do you have an answer to the above?
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    (Original post by NoFunnyBusiness)
    :confused: :rolleyes: Do you have an answer to the above?
    Yes, look at the mark scheme, it think it explains it pretty good.
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    (Original post by albus)
    Yes, look at the mark scheme, it think it explains it pretty good.
    Obviously the markscheme doesn't make sense to me if I posted for help...replies like this are just frustrating
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    What did you get for the equation of the line?
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    Go to this link, the person is pretty good
    http://bit.ly/A2Vector
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    (Original post by geditor)
    What did you get for the equation of the line?
    YAY

    I got r = (-1, -7, 11) + lambda (15, -20, 0)
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    DOES ANYONE DO MEI M2?
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    (Original post by NoFunnyBusiness)
    YAY

    I got r = (-1, -7, 11) + lambda (15, -20, 0)
    That's what I got How did you solve for lambda?
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    (Original post by geditor)
    That's what I got How did you solve for lambda?
    That's what I'm stuck and the markscheme isn't really explanatory
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    (Original post by albus)
    Go to this link, the person is pretty good
    http://bit.ly/A2Vector
    Am I meant to download something for this?
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    (Original post by NoFunnyBusiness)
    Am I meant to download something for this?
    no i'm pretty sure you don't
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    AE and BD are parallel, so have the same direction vector.

    From part i), you can see the vector AE is [15, -20, 0], so the position vector of E is (after taking '5' out the vector as lambda)
    r= [0, 0, 6] + (lambda)[3, -4, 0], where lambda=5

    You know the length AE is 25 from part i), and you know the length BD is 15 (given), so you can replace lambda=5 by lambda =3 for the position vector of D. From there, just replace [0,0,6] with [-1,-7,11] and solve for D.


    (those should all be column vectors, btw )
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    (Original post by Xenite)
    AE and BD are parallel, so have the same direction vector.

    From part i), you can see the vector AE is [15, -20, 0], so the position vector of E is (after taking '5' out the vector as lambda)
    r= [0, 0, 6] + (lambda)[3, -4, 0], where lambda=5

    You know the length AE is 25 from part i), and you know the length BD is 15 (given), so you can replace lambda=5 by lambda =3 for the position vector of D. From there, just replace [0,0,6] with [-1,-7,11] and solve for D.


    (those should all be column vectors, btw )

    Why can you replace lambda = 5 by lambda = 3, How did you arrive at knowing lambda =5 in the first place

    (I'm probably asking a stupid question)
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    You know that to get from A to E you start from A, and travel [15,-20,0] to get to E.

    As AE is a line, you can add in a variable lambda to the equation to get any point along that line or, if you change the start point (from A to B) to get any point on any parallel line (such as D)

    The actual equation for A to E specifically is [0,0,6]+[15,-20,0]. You can take out a multiple of 5 to make it [0,0,6]+5[3,-4,0].

    The [3,-4,0] is the direction vector. 5 multiples of it will allow you to travel the 25 metres from A to E. By this logic, you need 3 multiples of it (lambda=3) to travel 15 metres; the distance BD.

    Simply combine this knowledge of BD=3[3,-4,0] with your start-point (B) and you'll arrive at D.


    Edit: It wasn't a stupid question; vectors are hard concepts to get. I take further maths aswell, and one of our modules is entirely upon vectors, but I still struggle with the C4 stuff. Some people find it easy to visualise, others prefer other kinds of math, such as mechanics or statistics.
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    (Original post by Xenite)
    You know that to get from A to E you start from A, and travel [15,-20,0] to get to E.

    As AE is a line, you can add in a variable lambda to the equation to get any point along that line or, if you change the start point (from A to B) to get any point on any parallel line (such as D)

    The actual equation for A to E specifically is [0,0,6]+[15,-20,0]. You can take out a multiple of 5 to make it [0,0,6]+5[3,-4,0].

    The [3,-4,0] is the direction vector. 5 multiples of it will allow you to travel the 25 metres from A to E. By this logic, you need 3 multiples of it (lambda=3) to travel 15 metres; the distance BD.

    Simply combine this knowledge of BD=3[3,-4,0] with your start-point (B) and you'll arrive at D.


    Edit: It wasn't a stupid question; vectors are hard concepts to get. I take further maths aswell, and one of our modules is entirely upon vectors, but I still struggle with the C4 stuff. Some people find it easy to visualise, others prefer other kinds of math, such as mechanics or statistics.

    THANK YOU SO MUCH, that made perfect sense! Yeah I much prefer Mechanics :P
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    (Original post by NoFunnyBusiness)
    THANK YOU SO MUCH, that made perfect sense! Yeah I much prefer Mechanics :P
    Ha, no problem! I'm the same, Mechanics just seems so much easier than this

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