Projecting a Vector onto a Plane

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Niles
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Homework Statement


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)?
 
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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].
 
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HallsofIvy said:
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].

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.
 
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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]
 
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OK, so just a generalization from 2D to 3D. Thanks for your help and time