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|a+b| = |a| + |b| implies a and b parallel?

  1. Jul 30, 2011 #1
    Is the following theorem true:

    Theorem: Suppose [itex]a, \, b \in \mathbb{R}^k[/itex]. If [itex] |a| + |b| = |a + b| [/itex], then [itex] |a| [/itex] and [itex] |b| [/itex] are parallel to each other in the same direction.

    I proved the converse, but I couldn't prove the theorem above. Please post the proof or the disproof of it, or a link of them. Thanks.
  2. jcsd
  3. Jul 30, 2011 #2
    Hi julypraise! :smile:

    The trick to proving this is to calculate the inproduct


    in two different ways. The first way involves


    The other way starts as


    Can you finish it?
  4. Jul 30, 2011 #3
    Thank you so much, micromass.

    So, by using the trick, I derived that

    [itex] 2|a||b| = 2<a, \, b> [/itex], and by the definition of inproduct

    [itex] |a||b| = |a||b|cos\theta [/itex]

    and therefore, [itex] \theta = 0 [/itex]

    where [itex] \theta [/itex] is the angle between.
  5. Aug 4, 2011 #4
    If `a` and `b` are parallel then āˆƒcāˆˆā„ s.t., a=cb. So then, ||a+b||=||cb+b||=||b||(|c+1|)ā‰ ||a||+||b||. Since ||a||=||cb||=(|c|)||b||.
    Last edited: Aug 4, 2011
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