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Let a and b denote 2 2d vectors

  1. Oct 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Let a and b denote two 2D vectors a = <a1, a2> b=<b1, b2>

    show directly that a . b ≤ ||a||||b||

    2. Relevant equations



    3. The attempt at a solution

    Im looking for a place to start here. Should i start by computing the dot product of a and b and then also finding the magnitude of a and b?
     
    Last edited: Oct 10, 2013
  2. jcsd
  3. Oct 10, 2013 #2

    LCKurtz

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    Are you asking for our permission?
     
  4. Oct 10, 2013 #3
    no, a push in the right direction.
     
  5. Oct 10, 2013 #4

    LCKurtz

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    Well, why don't you be brave and try your own suggestion?
     
  6. Oct 10, 2013 #5
    i did, " Should i start by computing the dot product of a and b and then also finding the magnitude of a and b"
     
  7. Oct 10, 2013 #6

    LCKurtz

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    So do it. Show us what you get when you calculate both sides. Can you tell if the inequality is true? Can you work on it so you can tell? Let's see some effort.
     
  8. Oct 10, 2013 #7
    so after doing it i got a1b1 + a2b2 ≤ (a1+a2)(b1+b2)

    which can be expanded out to a1b1 + a2b2 ≤ a1b1 + a1b2 + a2b1 + a2b2
     
    Last edited: Oct 10, 2013
  9. Oct 10, 2013 #8
    since theres more terms on the right side can it be said that its greater than the left side?
     
  10. Oct 10, 2013 #9

    LCKurtz

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    No. The first thing you need to do is use the correct formulas for ##\|a\|## and ##\|b\|##.
     
  11. Oct 10, 2013 #10

    Office_Shredder

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    ||a|| is not equal to (a1+a2).
     
  12. Oct 10, 2013 #11
    isnt ||a|| = Sqrt(a12 + a22)?
     
  13. Oct 10, 2013 #12

    LCKurtz

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    Yes, but that is not equal to ##a_1+a_2##.
     
  14. Oct 10, 2013 #13
    if theyre both squared cant you just take the sqrt and end up with a1+a2?
     
  15. Oct 10, 2013 #14

    LCKurtz

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    Does ##\sqrt{3^2+4^2} = 3+4##?
     
  16. Oct 10, 2013 #15
    ahh i see

    so then i get

    a1b1 + a2b2 ≤ √(a12+a22) + √(b12+b22)
     
    Last edited: Oct 10, 2013
  17. Oct 10, 2013 #16

    LCKurtz

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    Where did that + come from? Fix that and then respond to post #6.
     
  18. Oct 10, 2013 #17
    my mistake

    a1b1 + a2b2 ≤ √(a12+a22) √(b12+b22)

    how can i determine the truth of the inequality?
     
    Last edited: Oct 10, 2013
  19. Oct 10, 2013 #18

    LCKurtz

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    Work on it. How might you get rid of the square roots?
     
  20. Oct 10, 2013 #19
    (a1b1)2 + (a2b2)2 ≤ (a12+a22) + (b12+b22)

    I can then multiply out the right side which gives me something similar to what i got in step 7 just the a and b terms are all squared

    can i then subtract the left side to get 0 <= a12b22 + a22b12?
     
    Last edited: Oct 10, 2013
  21. Oct 10, 2013 #20

    LCKurtz

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    You have that bogus + sign in there again and if you are squaring both sides, your algebra is wrong. You are never going to get this problem correct if you can't do algebra correctly.
     
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