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Vectors i, j, k

  1. Sep 10, 2010 #1
    Hello,

    I am studying for a test I have on Monday and in my class we briefly went over adding and subtracting vectors in the y, x, and z components or in other words the i, j, and k.

    The problem is I do not understand how to add and subtract problems like these to find the length:

    32j + 30 k

    or 16i-32j-25k

    I looked through my books and notes and I can't figure it out. I also don't know how to find the length of something like

    45i

    Thanks for helping me out its driving me crazy trying to make sense of these things.
     
  2. jcsd
  3. Sep 10, 2010 #2

    berkeman

    User Avatar

    Staff: Mentor

    Mostly it's using the Pythagorean theorem to find the magnitude, and a little trig to convert between "rectangular" or "Cartesean" (i,j,k) coordinates and Magnitude/Direction coordinates.

    http://en.wikipedia.org/wiki/Polar_coordinate_system

    http://en.wikipedia.org/wiki/Rectangular_coordinates

    The length (magnitude) of 45i is 45, and the direction is pointing in the +x axis direction.

    For other vectors that several rectangular components, use the conversion techniques described in the wikipedia article. Hope that helps.
     
  4. Sep 11, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The whole point of using "i", "j", and "k" is that addition and subtraction can be done "component wise". That is (ai+ bj+ ck)+ (ui+ vj+ wk)= (a+ u)i+ (b+ v)j+ (c+ w)k.
    In particular, 32j+ 30k (the coefficent of i is 0) added to 16i- 32j- 25k is (0+ 16)i+ (32-32)j+ (30- 25)k= 16i+ 0j+ 5k= 16i+ 5k.

    The length of the vector xi+ yj+ zk is [itex]\sqrt{x^2+ y^2+ z^2}[/itex]
    In particular, the length of just xi is [itex]\sqrt{x^2}= |x|[/itex].
    The length of 16i- 32j+ 25 k is [itex]\sqrt{16^2+ (-32)^2+ 25^2}= \sqrt{256+ 1024+ 625}= \sqrt{1905}[/itex] which is about 43.6. The length of 45j is |45|= 45.
     
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