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Tricky Vector Problem

  1. Sep 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Two teams of explorers leave a common point. The first team travels 5 km north across a plain, then follows a river 15 degrees west of north for 7 km before making camp for the night. THe second climbs a ridge, traveling basically due northeast for 6 km along a trail that climbs upward to an altitude of 2.5 km. This team then follows the ridge northward at approximately 4 km before making camp for the night. Can the two teams communicate that night if their radios have a range of 5 km?
  2. jcsd
  3. Sep 2, 2009 #2


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    Show your attempted solution.
  4. Sep 2, 2009 #3
    Draw a picture showing the paths of both groups. Once you have the picture drawn and distances labled it should be easy.

    And yeah, you're suppossed to show relevant equations and your attempted solution.
  5. Sep 2, 2009 #4


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  6. Sep 2, 2009 #5
    I thought the best course may be to try and find the magnitude of each of the respective teams. So for team one I assigned [5x, 7y, 0z] and for team two [6x, 2.5y, 4z].
    If my calculations were right their respective magnitudes should be 8.60233 and 7.63217.

    From there I get a bit fuzzy.

    P.S. Sorry for not using the correct symbology with my explanation. I have no idea how to illustrate it on computer.
  7. Sep 2, 2009 #6
    You have the coordinates of two points in space just find the distance between the two.
  8. Sep 2, 2009 #7
    Right. But that is my question. Nothing is coming to mind that I can use without the help of some other factor (i.e. an angle).
  9. Sep 2, 2009 #8
    Unless I consider these points two more vectors and repeat the process...?
  10. Sep 2, 2009 #9
    There would be two ways you could do this, essentially the same. You've got a formula, one that you learned a long time ago, but probably haven't used in awhile. It's the distance formula which finds the distance between two points. Or: when you have two values, a and b, and need to find the distance between them you just subtract them.
  11. Sep 2, 2009 #10
    So in essence you just take the difference from the respective magnitudes?
  12. Sep 2, 2009 #11
    These are vectors, you add them via their components.
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