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Orthogonal projections

  1. Jul 4, 2007 #1
    Given:

    [tex]\vec A \cdot \vec B = non zero[/tex]
    and
    [tex]\theta[/tex] does not equal 0

    I can't seem to prove that Vector B minus the Projection of B onto A makes the orthogonal projection of B onto A.

    Can you help?
     
    Last edited: Jul 4, 2007
  2. jcsd
  3. Jul 4, 2007 #2
    I am not sure I understand your question properly. The first statement in fact implies the second statement in your question.
    You can see the proof by simple visualization, using laws of vector geometry of course.
    Alternatively, prove that the dot product of "Vector B - Projection of B onto A and vector A" is 0. Just put that statement in algebra and simplify.
     
  4. Jul 5, 2007 #3

    HallsofIvy

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    Science Advisor

    The projection of B onto A is
    [tex]\frac{\vec{A}\cdot\vec{B}}{|\vec{A}|^2}\vec{A}[/tex]
    so
    [tex](\vec{B}- proj_{\vec{B}}(\vec{B}))\cdot \vec{A}= \vec{A}\cdot\vect{B}- \frac{\vec{A}\cdot\vec{B}}{|\vec{A}|^2}\vec{A}\cdot\vec{A}[/tex]
     
  5. Jul 5, 2007 #4
    thank you halls of ivy.

    thanks hall of ivy
     
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