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Geometric vectors question

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Find a and b such that the point (a,-5,b) lies on the line passing through (-3,-1,3) and (3,-4,9)


    2. Relevant equations
    (maybe?) Component = P1P2 = (x2 - x1, y2 - y1)


    3. The attempt at a solution

    Since it's a line passing through, I thought I could just add the two points together which gave me (0, -5, 12), but I did not get the right answer. So I thought maybe you have to get the component which would essentially be the terminal points of the 2 points which would form a new vector?

    P1P2 = (x2 - x1, y2 - y1)

    but then I'm not sure which would be the point to use for P2 and P1.

    We just started vectors like last week, and I'm still pretty newbie with this stuff :( I'm not totally sure how to interpret the question.

    What relevance does the -5 in (a, -5, b) have? It matches the addition of both of the points, but 0 and 12 do not work.

    Thanks for any help.
     
  2. jcsd
  3. Oct 27, 2009 #2

    Mark44

    Staff: Mentor

    You are given points in R3, but you're thinking in two dimensions. E.g.
    Notice that your points have three coordinates, not two!

    Lets's give some names to your points: A(-3,-1,3) and B(3,-4,9) and C(a,-5,b) . You want the vector AB to be parallel to the vector AC, which means that AB = t*AC for some real number t.

    As it turns out, the point C can't be between points A and B, but it should be on the line through them.
     
  4. Oct 27, 2009 #3
    How do you know they want AC and not BC? I still don't understand the wording of the question? How do you know it's supposed to be parallel? Is it because if it passes both points it means that AC contains the line AB?

    I'm still not sure on how to interpret this problem.

    AB = (-3,-1,3) - (3,-4,9) = (-6, 3, -6)

    AC = (a+3, -4, b-3)

    So now I need to figure out a scalar multiple that makes -4 into 3 which would be [itex]-\frac{3}{4}[/itex].

    Am I getting anywhere with this question? How do you know it's AB and now BA? It wouldn't be the same, would it? Sorry, I'm just not understanding this... matrices was way easier than vectors... Appreciate the help.
     
  5. Oct 27, 2009 #4

    Mark44

    Staff: Mentor

    You could do it that way. I just chose one way.
    AC has to have the same direction as the line through A and B. For that to happen, the direction vectors have to be scalar multiples of one another. I didn't mean parallel in the sense of two lines a certain distance apart. A better choice of wording might have been "collinear."

    AB is not the line. AB is the vector from A to B, so it has a certain length and a certain direction, the same as the line through A and B. The line has infinite length.
    Find a and b so that the point C(a, -5, b) is on the line that passes through the points A and B. IOW, Find C so that AC and BC have the same direction.
    That works, since (-3/4)*(-4) = 3
    It doesn't matter if it's AB or BA. While not equal, these vectors have directions that are opposite in sign. IOW, AB = -1*BA.
     
  6. Oct 27, 2009 #5
    So (-6, 3, -6) divided by -3/4 = (8, -4, 8), which means I have to make a and b add up to 8.

    AC = (a+3, -4, b-3)

    a = 5, b = 11

    I think I need to review the theory a bit more, but I got this question down. It's frustrating because you spend your whole life dealing with x and y axis and all of a sudden, BAM, here's a z-axis for ya to confuse the hell out of you. I tried drawing out the graphs for visual aide, but after plotting it out, I can't make any sense of it cuz my drawings don't look 3d.

    Anyways, thanks a lot for the help Mark, really appreciate it!
     
  7. Oct 27, 2009 #6

    Mark44

    Staff: Mentor

    Those are the right values. Using them, you can see that AC = (8, -4, 8) and BC = (2, -1, 2), so each vector is some multiple of the other, which makes them parallel - just what we wanted.

    For this problem, you don't really need to graph things in 3D. All three points are supposed to be in a line, so any plane that contains this line is just fine.

    In general, though, you'll want to learn how to graph things in three dimensions. Hopefully your instructor will start doing that soon, especially now that you are working with points in 3D space.
     
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