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Calc 3, vector problem

  1. Sep 19, 2011 #1
    I mostly just want to know if I did this correctly... And if not, where I went wrong.

    1. The problem statement, all variables and given/known data
    Of [itex]\overline{U}[/itex]=(-2, 6) is the vector [itex]\overline{AB}[/itex] and the midpoint of the line segment from A to B has the coordinates (2,1). Find A and B.
    [itex]\overline{A}[/itex]=(a1, a2)
    [itex]\overline{B}[/itex]=(b1, b2)

    2. Relevant equations

    3. The attempt at a solution
    [itex]\overline{AB}[/itex]=(a1, b1) + t(b1-a1, b2-a2)

    rearranging I got:
    [itex]\overline{AB}[/itex]=(1-t)(a1, b1) + t(a2, b2)

    So when t=0 I have the coordinates (a1 and b1)
    and when t=1 I have the coordinates (a2 and b2)

    So plugging the numbers:
    [itex]\overline{AB}[/itex]=(-2,6) + t(-2 - 2, 6 - 1)
    =(1-t)(-2, 6) + t(2, 1)
    [itex]\overline{A}[/itex]=(-2, 2)
    [itex]\overline{B}[/itex]=(6, 1)

    Did I do it right? or am I way off?... lol
  2. jcsd
  3. Sep 19, 2011 #2

    i think something is wrong. if t=0 , you should get point A, the coordinate
    [itex](a_1,b_1)[/itex] is not point A.

    I will do this as following. let O be the origin. and let the coordinates of A and B be
    (a1,a2) , (b1,b2). let the M be midpoint of
    the segment AB , then




    now realize that vectors AM and MB are the same since they have same magnitude and direction.


    so you get 4 equations and 4 unknowns . solve it. I got

    [tex]A=(3,-2) \; , B=(1,4)[/tex]

    just to confirm the answer, use midpoint formula for A and B and you get back the coordinate of point M.
  4. Sep 19, 2011 #3
    Thank you very much! You're awesome! :)
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