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Homework Help: Magnitude of 3D vector

  1. Dec 7, 2004 #1
    If I'm given a vector in 3D and it asks me to find the magnitude, I just take the square root of the numbers squared. But if I'm just given a magnitude and direction how could I find the vector?
    Last edited: Dec 7, 2004
  2. jcsd
  3. Dec 7, 2004 #2


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    If u have the magnitude and the direction,then U HAVE THE VECTOR.U're probably asking how to determine the components along the (presumably reciprocal orthogonal) axis of coordinates if u'regiven the mangitude and direction.It all stays in the the word "direction" that would require someangles between the vector's direction and the coordinate axis/planes,perhaps.From here on,there's nothing that plain space geometry.
  4. Dec 8, 2004 #3
    Suppose a vector has magnitude 8 and its direction is given to be

    [tex]\frac{1}{\sqrt{3}}i + \frac{1}{\sqrt{3}}j + \frac{1}{\sqrt{3}}k [/tex]

    then the vector u need is

    [tex] \frac{8}{\sqrt{3}}i + \frac{8}{\sqrt{3}}j + \frac{8}{\sqrt{3}}k[/tex]

    This should help you! :biggrin:
    Last edited: Dec 8, 2004
  5. Dec 8, 2004 #4


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    The magnitude is a number of course, but how are you given the directions? If you are given a unit vector in that direction, just multiply by the magnitude as prasanna said.

    If you are given the angles the vector makes with the coordinate axes, you can use the "direction cosines": if the angle the direction makes with the x axis is u, with the y axis v, and with the z axis w then cos(u)i+ cos(v)j+ cos(w)k is a unit vector in that direction. Multiply that by the magnitude.
  6. Dec 8, 2004 #5


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    The other format you might see is a magnitude and two angles. Usually one of the angles projects the vector onto the fundamental plane (latitude of some location, for example). The other gives the angle of projection relative to the x-axis (longitude of some location, for example).

    (Probably not a great example unless you realize it's just assumed that everyone knows the radius of the Earth).

    In this case, it would be:

    [tex]r cos (lat) cos (long)_i + r cos (lat) sin (long)_j + r sin (lat)_k[/tex]
    Last edited: Dec 8, 2004
  7. Dec 8, 2004 #6
    I was given a magnitude (a number) and the direction is the same as this vector (and the vector was given as ?i + ?j +?k).
    So I think I can go from here. I appreciate the input from all. :smile:
  8. Dec 9, 2004 #7
    Someone argued that a zero vector has direction. Isn't this so ridiculous? As far as I know a zero vector is just a point. How can a point have directions?
  9. Dec 10, 2004 #8
    A zero vector is a point which can have any direction!.
    I remember reading something like this
  10. Dec 10, 2004 #9

    [tex] \hat{a} = \frac{a}{|a|} [/tex]


    [tex]a = \hat{a}*|a|[/tex]

    simple algebra ;)
  11. Dec 10, 2004 #10
    I presume you're replying to jjiimmy101's original question on top? :uhh:

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