Calculating Orthogonal Projection: Proving AD and Formula Progress | \pi\rangle

In summary, the conversation revolves around proving the formula for AD using the concept of orthogonal projection. The main focus is on simplifying the right-hand side of the formula using the dot product. The conversation also touches on the concept of normalizing vectors and how it relates to the formula. Overall, the conversation provides a clear understanding of the steps involved in proving the formula for AD.
  • #1
Petrus
702
0
Hello MHB,
I have hard to prove that AD, I did put on pic the formula for AD and my progress is at bottom.
35l7rzd.png


Regards,
\(\displaystyle |\pi\rangle\)
 
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  • #2
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?
 
  • #3
Re: orthogonal projection

Ackbach said:
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,
\(\displaystyle |\pi\rangle\)
 
  • #4
Re: orthogonal projection

Petrus said:
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?
 
  • #5
Re: orthogonal projection

Ackbach said:
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,
\(\displaystyle |\pi\rangle\)
 
  • #6
Re: orthogonal projection

Petrus said:
\(\displaystyle u*v=|u||v|\cos\theta\)

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?
 
  • #7
Re: orthogonal projection

Ackbach said:
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,
\(\displaystyle |\pi\rangle\)
 
  • #8
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?
 
  • #9
Re: orthogonal projection

Ackbach said:
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
\(\displaystyle |\pi\rangle\)
 
  • #10
Re: orthogonal projection

Petrus said:
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.
 
  • #11
Re: orthogonal projection

Ackbach said:
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,
\(\displaystyle |\pi\rangle\)
 

Related to Calculating Orthogonal Projection: Proving AD and Formula Progress | \pi\rangle

What is orthogonal projection?

Orthogonal projection is a mathematical concept used to represent a higher-dimensional object onto a lower-dimensional subspace while preserving its shape and orientation. In simpler terms, it is a way of creating a 2D or 3D representation of a 4D or higher-dimensional object.

What is the purpose of orthogonal projection?

The purpose of orthogonal projection is to simplify complex objects by reducing their dimensionality while retaining important geometric properties. This allows for easier visualization and analysis of the object.

What are some common applications of orthogonal projection?

Orthogonal projection is commonly used in computer graphics, engineering, physics, and other fields where complex objects need to be represented in a simplified manner. It is also used in data analysis and machine learning to reduce the dimensionality of large datasets.

What is the difference between orthogonal projection and oblique projection?

Orthogonal projection is a type of projection that preserves angles and distances, while oblique projection does not. In oblique projection, the object is projected onto the viewing plane at an angle, resulting in distorted shapes and sizes. Orthogonal projection, on the other hand, maintains the original proportions of the object.

How is orthogonal projection calculated?

The calculation of orthogonal projection involves finding the dot product between the vector representing the object and the unit vectors of the projection plane. The result is then multiplied by the unit vectors to obtain the coordinates of the projected object. This process is repeated for each point of the object to create the final projection.

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