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Vector Magnitude Proof (Geometrically)

  1. Jan 19, 2006 #1
    I have this question from my book:

    Prove that |A-B| >= |A| - |B|

    Note that A and B are vectors, and that |A| means magnitude of vector A.
    ------

    I have already proved this using a cheap trick, but I want to prove it using geometry instead. My question is about what cases I would need. Meaning that since there is a triangle formed (if I put A, B, and A-B together) what would I need to do in order to provide a sufficient geometric proof.

    I was thinking of having A - B form a triangle and having a triangle with an acute angle between A and B, and then having a triangle with an obtuse angle between A and B, and then having A and B form a straight line. And of course showing on each of these that |A-B| >= |A| - |B|

    Would that be sufficient?

    Thanks.
     
  2. jcsd
  3. Jan 19, 2006 #2

    robphy

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    Do you recognize the inequality?

    By "geometrically", are algebraic vector methods or trigonometric methods allowed? Or are you looking for a "visual proof without words"?

    Your method looks sufficient to me (assuming the zero angle and straight angles are treated)... but possibly not elegant... unless, for example, the obtuse case uses the acute case.

    my $0.01
     
  4. Jan 20, 2006 #3

    0rthodontist

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    You can visualize the set of all vectors of length |a| as a circle and the set of all vectors of length |b| as another circle enclosing or being enclosed by the first. Then any line segment drawn between the two circles obviously has length greater than or equal to the line segments that go perpendicularly from the inner circle to the outer. That means |a - b| >= ||a| - |b|| so |a - b| >= |a| - |b|.
     
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