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Vectors in three dimensions

  1. Sep 14, 2008 #1
    I am trying to teach myself AS maths. This is from Introducing mechanics a txt book by Jefferson and Beadsworth. exercise 2C question 1 part vi)

    Question given vector b=6i-3j-2k;
    |b| is established as = 7 by calculation

    Calculate the angles between b and positive x-, y- and z- directions.

    My answer: these are
    sine alpha =6/7 = 59 degrees - subtract from 180 to give angle from z axis

    Sine beta =2/7 = 16.6 degrees - this is angle from x axis

    Sine theta = 3/7 = 25.4 degrees. Need to add 90 degrees to give angle from y axis.

    However these are not the correct answers. The correct answers appear to be cosines of the above. This does not make sense to me can you expalin?
  2. jcsd
  3. Sep 14, 2008 #2


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    Once you normalize the vector to unit length, the coeficients of (i,j,k) are the cosines with respect to the axes (6/7,-3/7,-2/7).
  4. Sep 14, 2008 #3


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    Welcome to PF!

    oh-oh-oh-enryiggins! Welcome to PF! :smile:
    Just you wait, enryiggins, just you wait.
    You'll be sorry, but your tears :cry: will be too late. :wink:
    The component in a direction is always multiplied by the cos, not the sin.

    After all, cos0º = 1, and if the angle is 0º, then the whole of the vector is the component in that direction … so it has to be cos, doesn't it? :wink:
  5. Oct 5, 2008 #4
    Many thanks for your help. My drawing was not correct hence the confusion.
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