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Linear Algebra: Projection Theortical Problem

  1. Oct 1, 2005 #1
    Hey Everyone,
    I have this question that's been giving me a hard time, I dont really know how to do it.

    "Let A be an arbitrary vector. It may be projected along a direction V on the plane P with normal vector n. What is its image A` ?"

    I know that A + lamda*V = A` , and that we have to do something with normal vector (perhaps a dot product with A` to eliminate lamda from the equation) to get a general formula for this situation, however I am unable to get an answer, any help would be greatly appreciated, thanks!

    A.Z.H
     
  2. jcsd
  3. Oct 1, 2005 #2
    [tex]\text{proj}_{\vec{v}}\vec{a}=\frac{\left(\vec{a},\vec{u}\right)}{\left(\vec{u},\vec{u}\right)}\,\vec{u}[/tex]

    ...where (x,y) is an inner product (a dot product in this case).
     
    Last edited: Oct 1, 2005
  4. Oct 1, 2005 #3
    That's not really what's being asked in the question, we dont want the projection of A onto V, we just want an a general expression for projecting A on the plane by just adding a multipe of V to it.
    Any ideas? :frown:
     
  5. Oct 1, 2005 #4
    The component (the lenth of the "shadow" cast upon the vector v) will be equal to |a|cosθ (just use trig for that). This can be written in a more convenient form like so:

    [tex]\text{comp}_{\vec{v}}\vec{a}=|a|\cos{\theta}=\frac{|a||v|\cos{\theta}}{|v|}=\frac{\vec{a}\cdot\vec{b}}{|v|}[/tex]

    ...now you want to find the component, which is just the vector with the above magnitude in the direction of v. To do this, just multiply the component by the unit vector in the direction of v, which is:

    [tex]\hat{v}=\frac{\vec{v}}{|v|}[/tex]

    [tex]\text{proj}_{\vec{v}}\vec{a}=\frac{\vec{a}\cdot\vec{v}}{|v|}\,\frac{\vec{v}}{|v|}=\frac{\vec{a}\cdot\vec{v}}{\vec{v}\cdot\vec{v}}\,\vec{v}[/tex]

    ...using the fact that [itex]|v|^2=\vec{v}\cdot\vec{v}[/itex].

    Do you understand?
     
  6. Oct 1, 2005 #5
    Since you changed your response since my last post:

    If v is in the plane P and A is not, there is no way (that I know of) to just add a multiple of v to A and have the resulting vector be in P. It doesn't make sense geometrically (that multiple would have to have an infinite magnitude).
     
  7. Oct 1, 2005 #6
    Hmmmm, yes you are right, if it is in the plane it's impossible.
    Im really sorry for all your trouble, I really appreciate your help!
    But assuming V is not in the plane, would it be possible? :smile:
     
  8. Oct 1, 2005 #7
    Well you can always find a plane which v would be in. I recommend just going with the definition of a projection on this one. Sorry I can't help any more, maybe someone else here can.
     
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