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C4 Vectors

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    I'm stuck on this Vectors question. It may be an easy question, but I really don't know what to do.

    Points A and B have coordinates (2,7) and (-3,3) respectively. Use a vector method to find the coordinates of C and D, where

    (a) C is the point such that AC= 3AB
    (b) D is the point such that AD= 3/5 AB


    What do I do? I don't want any solutions or anything, I just want to know what to do :/

    Thanks in advance
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    (Original post by Sem193)
    I'm stuck on this Vectors question. It may be an easy question, but I really don't know what to do.

    Points A and B have coordinates (2,7) and (-3,3) respectively. Use a vector method to find the coordinates of C and D, where

    (a) C is the point such that AC= 3AB
    (b) D is the point such that AD= 3/5 AB


    What do I do? I don't want any solutions or anything, I just want to know what to do :/

    Thanks in advance
    You should know that \vec{AB} = \mathbf{b} - \mathbf{a}
    Hence, you can calculate 3AB.

    Again, you should know that \vec{AC} = \mathbf{c} - \mathbf{a} , and if you equate 3AB and this it should be obvious how to work out the position vector of c.

    The second question is of very much the same idea.
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    (Original post by Sem193)
    I'm stuck on this Vectors question. It may be an easy question, but I really don't know what to do.

    Points A and B have coordinates (2,7) and (-3,3) respectively. Use a vector method to find the coordinates of C and D, where

    (a) C is the point such that AC= 3AB
    (b) D is the point such that AD= 3/5 AB


    What do I do? I don't want any solutions or anything, I just want to know what to do :/

    Thanks in advance
    Find AB, by using AB=OB-OA. OB and OA are the position vectors of B and A respectively.
    Then multiply the AB vector by 3 to find AC.

    Part (b) is similar to 'a' if you are able to do 'a' then you should be able to do 'b' as well.
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    I haven't done this chapter yet but this is what i would do:
    1) Find the distance between coordinates A and B using modulus formula.
    2) AC = 3AB so multiply your answer from part 1 by 3 to get AC
    3) Multiply answer in part 1 by 3/5 to get AD.
    4)Let the coordinates of C = (x,y)
    AC = distance from A to C.
    use distance formula to find AC using (x,y) and (2,7)
    the distance you get equals the answer you got from step 2 (3AB)
    solve simultaneously to get x and y values
    5) Follow a similar method to step 4 in order to find coordinates of D.

    That's what I would do but it may not be correct.
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    (Original post by SBarns)
    You should know that \vec{AB} = \mathbf{b} - \mathbf{a}
    Hence, you can calculate 3AB.

    Again, you should know that \vec{AC} = \mathbf{c} - \mathbf{a} , and if you equate 3AB and this it should be obvious how to work out the position vector of c.

    The second question is of very much the same idea.
    Thank you! I totally understand it now. I tried what you said and got the right answer. Thanks again!
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    You're welcome.
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    Thanks everyone who answered

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