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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

I have hard to prove that AD, I did put on pic the formula for AD and my progress is at bottom.

Regards,

\(\displaystyle |\pi\rangle\)

- Thread starter Petrus
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- Thread starter
- #1

- Feb 21, 2013

- 739

I have hard to prove that AD, I did put on pic the formula for AD and my progress is at bottom.

Regards,

\(\displaystyle |\pi\rangle\)

- Admin
- #2

- Jan 26, 2012

- 4,191

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?

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- #3

- Feb 21, 2013

- 739

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

Regards,

\(\displaystyle |\pi\rangle\)

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- #4

- Jan 26, 2012

- 4,191

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

Regards,

\(\displaystyle |\pi\rangle\)

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

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- #5

- Feb 21, 2013

- 739

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

Regards,

\(\displaystyle |\pi\rangle\)

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- #6

- Jan 26, 2012

- 4,191

Right. So why don't you use that to simplify the RHS, and see if you get the LHS? Then what could you do?\(\displaystyle u*v=|u||v|\cos\theta\)

Regards,

\(\displaystyle |\pi\rangle\)

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- #7

- Feb 21, 2013

- 739

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)Right. So why don't you use that to simplify the RHS, and see if you get the LHS? Then what could you do?

Regards,

\(\displaystyle |\pi\rangle\)

- Admin
- #8

- Jan 26, 2012

- 4,191

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?

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- #9

- Feb 21, 2013

- 739

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?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?

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

\(\displaystyle |\pi\rangle\)

- Admin
- #10

- Jan 26, 2012

- 4,191

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.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

\(\displaystyle |\pi\rangle\)

- Thread starter
- #11

- Feb 21, 2013

- 739

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

Regards,

\(\displaystyle |\pi\rangle\)