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Project vector onto a plane

  1. Jul 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi

    Say I have a vector v=(vx, vy, vz) that makes an angle α with the x-axis. The amount of v projected onto the x-axis I find from the dot product

    v.x = √(vx2 + vy2 + vz2)cos(α)

    I am also interested in finding the amount of v in the plane normal to the x-axis, i.e. I want to project v onto the y-z plane. Is this simply given by v.(y + z)?
     
  2. jcsd
  3. Jul 7, 2013 #2

    HallsofIvy

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    The components of a vector, v, at angle [itex]\theta[/itex] to a line, parallel and perpendicular to the line, form the legs of a right triangle having the vector as hypotenuse. Yes, the component parallel to the line is [itex]|v|cos(\theta)[/itex]. And just as obviously, the component perpendicular to the line is [itex]|v|sin(\theta))[/itex].
     
  4. Jul 7, 2013 #3
    Thanks for your reply. OK, I took an example to test this out. My (unit) vector is [itex]v = (0, v_y, v_z)/\sqrt{2}[/itex] and the vector I want to project v onto is
    [tex]
    u = (u_x, u_y, 0)
    [/tex]
    The projection v onto u is then (disregarding the √2...)
    [tex]
    u_y = \sqrt{u_x^2 + u_y^2}\cos \theta
    [/tex]
    The perpendicular part is then given by [itex]\sin\theta[/itex], which must be given by (since sin2 + cos2 = 1...)
    [tex]
    u_x = \sqrt{u_x^2 + u_y^2}\sin \theta
    [/tex]


    There is something I can't figure out - how can it be that there is no reference to any z-component in these expressions? I mean the above (perpendicular) sin-part only accounts for how much of vy is perpendicular to u, it does not include the fact that vz is also perpendicular. Is this something I have to manually add somehow? I hope you understand my question.
     
    Last edited: Jul 7, 2013
  5. Jul 8, 2013 #4

    HallsofIvy

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    Well, I avoided mentioning components at all for exactly that reason. if [itex]\vec{u}= u_x\vec{ i}+ u_y\vec{j}+ u_z\vec{k}[/itex] then its projection onto any line with which it makes angle [itex]\theta[/itex] is [itex]|u|cos(\theta)= \sqrt{u_x^2+ u_y^2+ u_z^2} cos(\theta)[/itex] and the length of the perpendicular is [itex]|u|sin(\theta)= \sqrt{u_x^2+ u_y^2+ u_z^2} sin(\theta)[/itex]
     
    Last edited: Jul 8, 2013
  6. Jul 9, 2013 #5
    OK, so just a generalization from 2D to 3D. Thanks for your help and time
     
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