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Orthogonal projection

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I have hard to prove that AD, I did put on pic the formula for AD and my progress is at bottom.


Regards,
\(\displaystyle |\pi\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: orthogonal projection

What is $\vec{u} \cdot \vec{v}$? I object to the notation $\cos(AC)$. You take the cosine of an angle, not the length of a side. Of which angle are you taking the cosine?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: orthogonal projection

What is $\vec{u} \cdot \vec{v}$? I object to the notation $\cos(AC)$. You take the cosine of an angle, not the length of a side. Of which angle are you taking the cosine?
ops i meant \(\displaystyle \cos\theta*AC\) Is my post unclear what I ask for?

Regards,
\(\displaystyle |\pi\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: orthogonal projection

ops i meant \(\displaystyle \cos\theta*AC\) Is my post unclear what I ask for?

Regards,
\(\displaystyle |\pi\rangle\)
No, the post is not unclear. I agree that what you're trying to prove is actually true. I was asking a Socratic question. You should write $AC \, \cos(\theta)$, so that's it's crystal clear what is inside the argument of the cosine, and what is not.

So, let me repeat the question: can you write out what $\vec{u} \cdot \vec{v}$ is?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: orthogonal projection

No, the post is not unclear. I agree that what you're trying to prove is actually true. I was asking a Socratic question. You should write $AC \, \cos(\theta)$, so that's it's crystal clear what is inside the argument of the cosine, and what is not.

So, let me repeat the question: can you write out what $\vec{u} \cdot \vec{v}$ is?
\(\displaystyle u*v=|u||v|\cos\theta\)

Regards,
\(\displaystyle |\pi\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: orthogonal projection

\(\displaystyle u*v=|u||v|\cos\theta\)

Regards,
\(\displaystyle |\pi\rangle\)
Right. So why don't you use that to simplify the RHS, and see if you get the LHS? Then what could you do?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: orthogonal projection

Right. So why don't you use that to simplify the RHS, and see if you get the LHS? Then what could you do?
Hmm... One quest, could I get to that formula without knowing RHS ( I was more planing to learn how to get to that formula insted of 'memorize' it in exam)

Regards,
\(\displaystyle |\pi\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: orthogonal projection

Ah, I see. Well, here's how I would approach it. Start with a vector in the correct direction, $\vec{u}$. Normalize it by dividing by its length: $\vec{u}/|u|$. Now I have a unit vector (that is, a vector of length $1$), in the direction that I want. The goal is to compute $AD$, and multiply my unit vector by this length $AD$. Now, we know that
$$AD=AC \, \cos( \theta).$$
You'd like to write the RHS using the dot product. So, add in whatever is missing. Can you see how to finish?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: orthogonal projection

Ah, I see. Well, here's how I would approach it. Start with a vector in the correct direction, $\vec{u}$. Normalize it by dividing by its length: $\vec{u}/|u|$. Now I have a unit vector (that is, a vector of length $1$), in the direction that I want. The goal is to compute $AD$, and multiply my unit vector by this length $AD$. Now, we know that
$$AD=AC \, \cos( \theta).$$
You'd like to write the RHS using the dot product. So, add in whatever is missing. Can you see how to finish?
those normalize formula I don't understand how we get it but I understand what it is. I understand exemple \(\displaystyle 1=\gamma u\) then \(\displaystyle \frac{1}{u}= \gamma\) that is so far I understand about but how do they do after?

so I got:
\(\displaystyle \frac{|v|cos\theta}{|u|}*u\) then multiplicate both side with |u| and use dot product to write the top as u*v?
is this correct?

Regards
\(\displaystyle |\pi\rangle\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: orthogonal projection

those normalize formula I don't understand how we get it but I understand what it is. I understand exemple \(\displaystyle 1=\gamma u\) then \(\displaystyle \frac{1}{u}= \gamma\) that is so far I understand about but how do they do after?

so I got:
\(\displaystyle \frac{|v|cos\theta}{|u|}*u\) then multiplicate both side with |u| and use dot product to write the top as u*v?
is this correct?

Regards
\(\displaystyle |\pi\rangle\)
Basically correct, although I wouldn't multiply both sides by $|u|$, but I would multiply the top and bottom by $|u|$. Then you're done.
 

Petrus

Well-known member
Feb 21, 2013
739
Re: orthogonal projection

Basically correct, although I wouldn't multiply both sides by $|u|$, but I would multiply the top and bottom by $|u|$. Then you're done.
Thanks for the help and taking your time! Now I see how you got that formula :) Heh i meant top and bottom (Cool)

Regards,
\(\displaystyle |\pi\rangle\)