A conceptual question on vectors

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Homework Help Overview

The discussion revolves around the properties of vector magnitudes, specifically whether the magnitude of the difference between two vectors can exceed the sum of their magnitudes. Participants explore various scenarios and mathematical principles related to vector addition and the triangle inequality.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants question the relationship between vector magnitudes and explore the implications of the triangle inequality. Some suggest drawing a triangle to visualize the problem, while others reference specific cases involving obtuse angles.

Discussion Status

The conversation includes differing viewpoints on the initial question, with some participants asserting that the magnitude of the difference cannot exceed the sum, while others challenge this notion by referencing specific cases. There is an ongoing exploration of the mathematical principles involved, though no consensus has been reached.

Contextual Notes

Some participants reference prior knowledge of the cosine law and triangle inequality, indicating a shared understanding of these concepts, while also expressing uncertainty about their application in this context.

Naeem
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Q. Can the magnitude of the difference of two vectors be ever be greater than the sum of the magnitudes of the two?

Any insights, or answers shall be appreciated...
Thanks
 
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Consider the case when the angle between the 2 vectors is in the interval [itex]\left(\frac{\pi}{2},\pi\right) [/tex]...What happens then...?<br /> <br /> Daniel.<br /> <br /> EDIT:Nothing happens.See posts #5 & #6.[/itex]
 
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The answer is no, but, of course, you should ask why!

Draw a triangle, with sides [itex]a, b, c[/itex] and with angles, opposite to the respective sides, [itex]A, B, C[/itex]. Now, remember way back in the past when you learned the cosine law? To remind you, it says that for such a triangle,

[tex]c^2 = a^2 + b^2 - 2ab\cos C.[/tex]

Now, see if that helps at all :smile:

Actually drawing the triangle is advisable!
 
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The answer is YES for obtuse triangles ([itex]C\in \left(\frac{\pi}{2},\pi\right)[/itex]) just as i suggested above...:wink:

Daniel.

EDIT:See posts #5 & #6.
 
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No it's not~

http://mathworld.wolfram.com/TriangleInequality.html

dexter forgetting things in your old age! :biggrin:

What we want is

[tex]\| x - y \| \leq \|x\| + \|y\|[/tex]

the triangle inequality is

[tex]\| x + y \| \leq \|x\| + \|y\|[/tex]

replace [itex]y[/itex] with [itex]-y[/itex] to obtain

[tex]\| x - y \| \leq \|x\| + \|-y\|[/tex]

but [itex]\|-y\| = \|y\|[/itex] so

[tex]\|x - y\| \leq \|x\| + \|y\|[/tex]

as we wanted.

I proved the Cauchy-Schwarz inequality, which leads to the triangle inequality, the other day:

https://www.physicsforums.com/showthread.php?t=69574
 
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I think people are forgetting everything lately... I've made several silly mistakes in the past few days. Probably all that air pollution these days :wink:
 
Bulls***.It's human nature.It reminds us that u're infallible...:wink:

Where would all the fun be,if everyone was perfect/right all the time...?

Daniel.

P.S.I know this post was a lame excuse...:-p
 
It reminds us that u're infallible... :wink:

I don't need to be reminded that I'm infallible! :-p
 
  • #10
Freudian slip??No,i sometimes forget the "no,not" all those negations...Of course i meant just the opposite...You see,errors make the fun in life...:-p

Daniel.
 
  • #11
A straight line is the shortest distance between 2 points: The sum of lengths of 2 sides of a triangle cannot be less than the length of the third side. That answers your question.
 

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