Proof of Orthogonal Projection (vectors)

runningeagle

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Oct 3, 2009
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14
Hi,

Let v and w be non zero vectors.
"Show that (w - projection of w onto v) is orthogonal to v."

I used geometry to show that this is true for an acute angle. I used the parallelogram method to prove it is orthogonal.

Is there a way to show orth[sub:1411rcl0]v[/sub:1411rcl0]w is orthogonal to v using algebra and definitions?
What would a vector projection look like geometrically if the angle were obtuse?

Thank you.
 
I started writing out a response before realizing that it is just wrong. The projection of w onto v is indeed parallel (consistent even) with v. The correct question might be "Show that (w - projection of w onto v) is orthogonal to v."
 
As far as your obtuse question, I'm not sure it makes sense. There would be no projection, unless the definition was made to project it onto the negative extension of the vector.

Algebraically you would need to show that the dot product was zero.
 
So, I am correct in only taking the acute case for my geometric proof?

How can I show algebraically that the dot product is zero?

I know, from the drawing, that proj[sub:dj3su541]v[/sub:dj3su541]w is Ai+0j and orth[sub:dj3su541]v[/sub:dj3su541]w is 0i+Aj and therefore the dot product of them is zero. I don't know how to solve algebraically without the geometric proof.

Thank you.
 
To be honest, working with orthonormal bases, I've only ever used algebra to verify orthogonality. I've never though about this problem.

The definition of the standard inner product on R^2 is: given u=(x,y), v=(x',y'), then <u,v> = (x * x') + (y * y').

The projection of w onto v is Proj_v(w) = [<w,v>/||v||^2]*v or ||w||/||v|| * cos(theta) * v

------ side note: from the second definition, you can see if theta is an angle between pi/2 and pi you get a negative value, which is the issue i mentioned in my last post.

You wish to show <w-Proj_v(w), v> = 0. You can assume the components of w are x,y and the components of v are (x',y')
 
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