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Projection of vectors

  1. Aug 24, 2005 #1
    I was copying my friends notes and had a hard time understanding one of the examples he had written down from lecture. See the attachment for a the picture of the example. This example looks like a projection of two vectors to me, but i'm not sure.

    [tex] u'=\frac{4i+2j}{\sqrt{20}} [/tex] u' = unit vector u in the direction of force
    [tex] v'=\frac{3i+4j}{5} [/tex] v' = unit vector v

    [tex] Fy' = 7kN (\frac{4i+2j}{\sqrt{20}})\cdot (\frac{3i+4j}{5})

    = 7kN (\frac{4}{\sqrt{20}} * \frac{3}{5} + \frac{2}{\sqrt{20}} * \frac{4}{5}) [/tex]

    The unit vectors came from the drawing and are in the direction of the two vectors. The thing i don't get is why are 2 unit vectors being dotted, then multiplied by the magnitude 7 kN?

    http://img394.imageshack.us/img394/5998/untitled3bz.png [Broken]

    **EDIT: forgot to include in the image a unit vector symbol in the pic for F, so F = 7 kN e' (e' is a unit vector)
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Aug 24, 2005 #2
    It is the same regardless of how you do it. If you multiply out the magnitude of the F vector onto the F's unit vector, you get the original vector F. You can then dot it against the y vector to find [itex] F \cdot y [/tex] (which I think is what your question is asking).
  4. Aug 25, 2005 #3


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    Homework Helper

    The subscript on the Fy should be v' instead! (ie, your e' = v')
    that is, your notes found the v' component of the 7kN Force
    which original vector was along the u' direction.

    F_y would be = F cos(phi), where cos(phi) = u'.j (<=dot product)
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