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Geometric proof for vector relation- !

  • Thread starter fahd
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geometric proof for vector relation-plz help!

hi there....
i am trying to prove the following relation from vectors geometrically however nothing comes to my mind..i have succeeded in proving it algebraically.
CAN ANYONE help me as to how do i prove this relation geometrically.

The relation is:

mag(A.B) <= mag A. mag B

where A and B are vectors
and mag stands for magnitude
 
Last edited:

Answers and Replies

  • #2
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fahd said:
hi there...
i am trying to prove the following relation from vectors geometrically however nothing comes to my mind..i have succeeded in proving it algebraically.
CAN ANYONE help me as to how do i prove this relation geometrically.

The relation is:

mag(A.B) <= mag A. mag B

where A and B are vectors
and mag stands for magnitude
you miss read the problem.....

It's the inner or dot product of two vectors is less than or equal to the magnitude of the two vectors multiplied together. This is pretty easy to do if you have some information.....

[tex]
cos(\theta) = \frac{A.B}{||A||*||B||}
[/tex]

Knowing the range of the cosine function allows you to say something about the fraction on the RHS of that equation....

Note that is the also known as the Cauchy inequality...
 
  • #3
40
0
Townsend said:
you miss read the problem.....

It's the inner or dot product of two vectors is less than or equal to the magnitude of the two vectors multiplied together. This is pretty easy to do if you have some information.....

[tex]
cos(\theta) = \frac{A.B}{||A||*||B||}
[/tex]

Knowing the range of the cosine function allows you to say something about the fraction on the RHS of that equation....

Note that is the also known as the Cauchy inequality...

hi there..
the proof that u have told me is the algebraic proof which i already know of..I used the cos limits being below 1 to prove that..BuT THE QUESTION IS HOW DO I PROVE THE PROBLEM GEOMETRICALLY????
 
  • #4
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fahd said:
hi there..
the proof that u have told me is the algebraic proof which i already know of..I used the cos limits being below 1 to prove that..BuT THE QUESTION IS HOW DO I PROVE THE PROBLEM GEOMETRICALLY????
You could make a geometric argument with the law of cosines but I don't know how good of a proof that really is. I mean....can that be extended beyond R^3?

What class is this for? I'm not sure I can offer you much help beyond suggesting that you use the law of cosines....

In any case the Cauchy bound is true in R^n and I don't know how to do geometric proofs in R^n or if they can even be done...
 
  • #5
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
17
Well, since you're only dealing with two vectors, you can always look at the plane they span, reducing it to a problem in R^2.

That being said, fahd, what are you using for the geometric meaning of the dot product?
 
  • #6
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this is my classical mechanics problem...there was another type of a question also better called the triangle inequality..which i geometrically proved by stating that the sum of sides of a triangle is always greater than the third side...however am confused as to what cud be a statement for this one to which i could relate a figure as well?>
 

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