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Let a and b denote 2 2d vectors

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  • #1
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Homework Statement



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

show directly that a . b ≤ ||a||||b||

Homework Equations





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?
 
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Answers and Replies

  • #2
LCKurtz
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Homework Statement



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

show directly that a . b ≤ ||a||||b||

Homework Equations





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?
Are you asking for our permission?
 
  • #3
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no, a push in the right direction.
 
  • #4
LCKurtz
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Well, why don't you be brave and try your own suggestion?
 
  • #5
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i did, " Should i start by computing the dot product of a and b and then also finding the magnitude of a and b"
 
  • #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.
 
  • #7
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so after doing it i got a1b1 + a2b2 ≤ (a1+a2)(b1+b2)

which can be expanded out to a1b1 + a2b2 ≤ a1b1 + a1b2 + a2b1 + a2b2
 
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  • #8
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since theres more terms on the right side can it be said that its greater than the left side?
 
  • #9
LCKurtz
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so after doing it i got a1b1 + a2b2 ≤ (a1+a2)(b1+b2)

which can be expanded out to a1b1 + a2b2 ≤ a1b1 + a1b2 + a2b1 + a2b2
since theres more terms on the right side can it be said that its greater than the left side?
No. The first thing you need to do is use the correct formulas for ##\|a\|## and ##\|b\|##.
 
  • #10
Office_Shredder
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||a|| is not equal to (a1+a2).
 
  • #11
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isnt ||a|| = Sqrt(a12 + a22)?
 
  • #12
LCKurtz
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isnt ||a|| = Sqrt(a12 + a22)?
Yes, but that is not equal to ##a_1+a_2##.
 
  • #13
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if theyre both squared cant you just take the sqrt and end up with a1+a2?
 
  • #14
LCKurtz
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Does ##\sqrt{3^2+4^2} = 3+4##?
 
  • #15
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ahh i see

so then i get

a1b1 + a2b2 ≤ √(a12+a22) + √(b12+b22)
 
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  • #16
LCKurtz
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ahh i see

so then i get

a1b1 + a2b2 ≤ √(a12+a22) + √(b12+b22)
Where did that + come from? Fix that and then respond to post #6.
 
  • #17
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my mistake

a1b1 + a2b2 ≤ √(a12+a22) √(b12+b22)

how can i determine the truth of the inequality?
 
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  • #18
LCKurtz
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my mistake

a1b1 + a2b2 ≤ √(a12+a22) √(b12+b22)

how can i determine the truth of the inequality?
Work on it. How might you get rid of the square roots?
 
  • #19
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(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?
 
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  • #20
LCKurtz
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(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?
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.
 
  • #21
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made a mistake and posted twice
 
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  • #22
Office_Shredder
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You're making the same error squaring the left hand side that you were making with taking square roots earlier.
 
  • #23
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so the left side should be (a1b1 + a2b2)2?
 
  • #24
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Yes.
It would be easier for us if you put each whole inequality so we don't have to go 10 or 15 posts back to see where you're coming from.
 
  • #25
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hopefully theres no algebra mistakes now

i started with
a1b1 + a2b2 ≤ √(a12+a22) + √(b12+b22)

squared both sides to get
(a1b1 + a2b2)2 ≤ (a12+a22) (b12+b22)

after working out both sides i end up with

2a1b1a2b2 <= a12b22 + a22b12
 
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