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Closest point to a vector

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Let A and B be given vectors in the plane R2. Find a formula for the closest point from A to the line along B.

    2. Relevant equations

    The idea is to derive an equation...

    3. The attempt at a solution

    The problem above is listed verbatim. I don't know how I would attempt it because I would just look at two vectors and say be able to see which point was closest. In other words, I have no idea how to derive a formula.
     
  2. jcsd
  3. Sep 11, 2011 #2

    lewando

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    "the closest point from A" seems a little vague. Why don't you assume the question is asking about the location of the "head" at (xa, ya) versus the tail at (0, 0) of A, a position vector.

    There is a method from algebra for finding a line (that goes through a specific point) that is perpendicular to another line. Refresh your menmory on that and try it.
     
  4. Sep 11, 2011 #3
    I have a question though. Can you/someone draw out what is going on?

    Wouldn't the answer always be the the tail (0,0), because presumably the other vector, B, is also starting at the origin. Thus the two vectors are starting on top of each other?
     
  5. Sep 11, 2011 #4

    lewando

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    Not to be difficult, but why don't you do it? It's pretty easy. Use Microsoft Paint and post your image on photobucket (or equivalent) then insert the link. It is a worthwhile exercise.

    I guess that's the easy way out. Heres the thing: a vector is not the same thing as a ray. A ray, loosely defined is half a line containing a tons of points. A vector is an entity possessing a magnitude (of something) and a direction, not so much "points". Vectors don't even need to be pinned down at the origin, like when you are doing head-to-tail vector addition. A displacement vector can also "float around" in space as long as the magnitude and direction information is preserved. For the sake of your question (which is sounding more ill-conceived the more I think about it), make the assumptions I suggested. Or better: talk with your teacher to seek clarification.
     
  6. Sep 11, 2011 #5
    Here is a picture of what, I think, is going on: http://imageshack.us/photo/my-images/535/vectordistance.png/

    I'm aware of what a vector is (took AP Calculus BC (5), and Physics B (5)), which is why I don't understand what this question is even getting at.
     
  7. Sep 11, 2011 #6

    lewando

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    Okay, nice work. Yes, this problem bugs me too. Is the placement of B your own, or is that specified by the problem?
     
  8. Sep 11, 2011 #7
    <img src=http://img535.imageshack.us/img535/6664/vectordistance.png> [Broken]
    That is my own drawing. I made up the placement of B.
     
    Last edited by a moderator: May 5, 2017
  9. Sep 11, 2011 #8

    lewando

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    I don't know what else to suggest. When you get one of these kinds of problems, you can do at least 2 things: 1) If you have time, seek clarification from the authorities. 2) If pressed for time, state your assumptions, and submit something. I would suggest assuming that both vectors are constrained to emanate from the origin. Your reference point for vector A is at the "head" (xA, yB). The line along vector B would then be y = mBx + 0. You need to commit one of these actions before you can really start this problem (please feel free to use your own set of assumptions, if you like).
     
  10. Sep 11, 2011 #9
    Banchoff...x_x
     
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