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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

    micromass

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    Hi julypraise! :smile:

    The trick to proving this is to calculate the inproduct

    [tex]<a+b,a+b>[/tex]

    in two different ways. The first way involves

    [tex]<a+b,a+b>=|a+b|^2=(|a|+|b|)^2[/tex]

    The other way starts as

    [tex]<a+b,a+b>=|a|^2+|b|^2+2<a,b>[/tex]

    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 ∃c∈ℝ s.t., a=cb. So then, ||a+b||=||cb+b||=||b||(|c+1|)≠||a||+||b||. Since ||a||=||cb||=(|c|)||b||.
     
    Last edited: Aug 4, 2011
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