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Linear Algebra Problem

  1. May 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Untitled.png

    Under what restrictions on ##c, d, e##, will the combinations ##c\vec u + d\vec v + e\vec w## fill in the dashed
    triangle?
    2. Relevant equations

    3. The attempt at a solution
    Clearly, ##\vec w + a (\vec v - \vec w) + b(\vec u - \vec v)## will be in the triangle when ##0 \leq b \leq a \leq 1## ;
    ##\vec w + a (\vec v - \vec w) + b(\vec u - \vec v) = b\vec u + (a-b) \vec v + (1-a)\vec w ##
    So, we have ##c=b, a-b=d, e=1-a##
    The restrictions are ,therefore ,
    ##c+d+e=1##
    ##0\leq c \leq 1,~ 0\leq d \leq 1,~ 0 \leq e\leq 1##
    But according to the book, the restrictions are, ##c+d+e=1;~ 0\leq c ,~ 0\leq d ,~ 0 \leq e;##
     
    Last edited: May 27, 2016
  2. jcsd
  3. May 27, 2016 #2

    SammyS

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    Gold Member

    You have a typo in that combination.

    It should be: ##c\,\vec u + d\,\vec v + e\,\vec w##
    Those answers are equivalent.

    If c, d, and e are all non-negative and if their sum does not exceed 1, then none of c, d, or e can exceed 1 .
     
  4. May 27, 2016 #3

    Ray Vickson

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    These say the same thing: the constraints ##c,d,e \geq 0##, ##c+d+e=1## imply ##c,d,e \leq 1## automatically (think about this). It is harmless but unnecessary to include the explicit upper bounds ##c,d,e \leq 1##.
     
  5. May 27, 2016 #4

    fresh_42

    Staff: Mentor

    If the sum equals 1 and all three summands are positive, how can one be bigger than 1?
     
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