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Homework Help: Projection of one vector on another?

  1. Dec 3, 2007 #1
    Projection of one vector on another??

    Can anyone explain how to find the projection of one vector along another?

    I thought it was scalar (dot) product, but then I realised it WASN'T. What is this then?

    Anyone explain?
  2. jcsd
  3. Dec 3, 2007 #2
  4. Dec 3, 2007 #3


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    projection of y onto x = x(x'x)-1x'y [= predicted value of y from the least squares equation "y = a + bx + u"].
  5. Dec 4, 2007 #4


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    The LENGTH of the projection of one vector onto another is (almost) the dot product.

    To find the projection of [itex]\vec{u}[/itex] on [itex]\vec{v}[/itex], draw the line from the "tip" of [itex]\vec{u}[/itex] perpendicular with [itex]\vec{v}[/itex]. You now have a right triangle with angle [itex]\theta[/itex] between the angles and hypotenuse of length [itex]|\vec{u}|[/itex]. The length of the projection, the "near side", is then [itex]|\vec{u}|cos(\theta)[/itex]. Since the dot product can be defined as [itex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/itex], to get the length of the pojection, we need to get rid of that [itex]|\vec{v}|[/itex] by dividing by it. The length of the projection of [itex]\vec{u}[/itex] on [itex]\vec{v}[/itex] is

    In order to get the projection vector itself, we need to multiply that length by the unit vector in the direction of [itex]\vec{v}[/itex], which is, of course, [itex]\vec{v}/|\vec{v}|[/itex].
    The vector projection of [itex]\vec{u}[/itex] on [itex]\vec{v}[/itex] is
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