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Vector problem in three dimensions

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data
    First of all, sorry if this isn't the right place to post this.

    Four points A, B,C and D have coordinates (0, 1, -2), (1, 3, 2), (4, 3, 4) and (5, -1, -2) respectively. Find the position vectors of
    a. The mid-point E of AC
    b. The point F on BC such that BF/FD = 1/3
    Use your answers to draw a sketch showing the relative positions of A, B, C and D



    2. Relevant equations



    3. The attempt at a solution
    I had no difficulty solving a. and b.: both E and F have position vectors (2i, 2j, k)
    This confuses as with respect to the last exercise, though. If E and F have the same position vectors, they should be the same point. Since E lies on AC and F on BC, shouldn’t A, B and C lie on the same line? However, vector AC = 4i + 2j + 6k and vector BC = 3i + 2k: they are not multiples, so A, B and C are not collinear.
    I’m new to the study of vectors in three dimensions so there’s probably a noob mistake I’m making somewhere. What am I missing?
     
  2. jcsd
  3. Jan 25, 2012 #2
    Take a look again at your midpoint for part (a). When finding the midpoint between A and C, we have ((4-0)/2, (3-1)/2, (4+2)/2). So the midpoint E doesn't have the coordinates you mentioned.

    Part (b), try finding the point F again: Let F = (a, b, c). We have 1/3 = BF/FD, which means that (1/3)FD = BF. Recall that FD = (5, -1, -2)-(a, b, c), and BF = (a, b, c)-(1, 3, 2).

    Then you will have a system of equations, and you will be able to find the coordinates of F.
     
  4. Jan 25, 2012 #3

    vela

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    You should write either (2, 2, 1) or ##2\hat{i}+2\hat{j}+\hat{k}##. Don't mix the notations.

    I agree that E is (2, 2, 1), but I get a much messier answer for F. How did you find F?
     
  5. Jan 25, 2012 #4

    vela

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    This isn't correct.

    This isn't correct either. BF and FD denote lengths.
     
  6. Jan 25, 2012 #5
    My apologies. I should try not to answer questions when I'm sleepy.
     
  7. Jan 26, 2012 #6
    Alright so here’s my work on b. :
    We have 3BF = FD, f = BF + b
    And since we know b, all we need is BF to find f position vector of F
    BF + FD = BD => 4BF = d – b = 4i – 4j – 4k => BF = i – j – k
    => f = (1+1)i + (3-1)j + (2-1)k = 2i + 2j + k

    Now I’m aware that my error lies in assuming the vector BF equals one-third of vector FD, as this proportion only applies to the lengths of the vectors and not the vectors themselves. It WOULD mean the same thing, however, if F lies on BD and not BC: so I’m starting to believe that problem is simply a typo in the textbook, in which case my working would be valid.
     
  8. Jan 26, 2012 #7

    vela

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    So do you still have the question about sketching the four points?
     
  9. Jan 27, 2012 #8
    yes! i'm not sure how to sketch a three-dimensional vector at all, especially if the points are not collinear.
     
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