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Homework Help: Projection Problem

  1. May 5, 2008 #1
    1. The problem statement, all variables and given/known data

    The projection of the vector V onto (a,b) = (a,b)
    The projection of the vector V onto (-b,a) = (-b,a)
    Describe V in terms of a and b

    2. Relevant equations



    3. The attempt at a solution

    I let V=(x,y) then place that into the projection equation for each to get:

    [(x,y)(a,b)/(a,b)(a,b)](a,b) = [ax+by/a^2+b^2](a,b)=(a,b)

    [(x,y)(-b,a)/(-b,a)(-b,a)](-b,a) = [-bx+ay/b^2+a^2](-b,a)=(-b,a)

    After seeing this I realized that:

    k(V)=V so k has to equal 1

    so both [ax+by/a^2+b^2] and [-bx+ay/b^2+a^2] should eqaul 1

    Then I let:
    [ax+by/a^2+b^2]=[-bx+ay/b^2+a^2]
    the denominators cancel each other out and I'm left with

    ax+by=-bx+ay


    After that I solved for y to get y=(a^2+b^2-ax)/b
    Subsituting that in I get:
    [-bx+a[(a^2+b^2-ax)/b]/b^2+a^2]

    Reducing and all that other good stuff I finally get x=[a^2+b^2-(a^3/b)+[(ab^2)/2]]/[(a^2)/b]-b]

    I am unaware if I am doing this correctly, maybe it's just how it looks that is throwing me off. Any advice or help would be greatly apperciated. THanks
     
    Last edited: May 5, 2008
  2. jcsd
  3. May 5, 2008 #2

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    You had two equations and two unknowns.

    This = 1 and That = 1.
    You only used one equation
    This = That.
    Now use either of the original.... they're linear so you could solve them via linear methods but its just as easy in this case to solve one eqn for y in terms of x and substitute it into the other. When you're done verify that both original equations are true.

    There is however a quicker way to do this problem... by noting that the two given vectors are orthogonal to each other!!!
     
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