Let a and b denote 2 2d vectors

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  • #1
ibaforsale
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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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  • #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
ibaforsale
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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
ibaforsale
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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
ibaforsale
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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
ibaforsale
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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
ibaforsale
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isnt ||a|| = Sqrt(a12 + a22)?
 
  • #13
ibaforsale
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if theyre both squared cant you just take the sqrt and end up with a1+a2?
 
  • #15
ibaforsale
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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
ibaforsale
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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
ibaforsale
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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
ibaforsale
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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
ibaforsale
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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
ibaforsale
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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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  • #26
LCKurtz
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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

Well yes, you finally have the algebra correct. So what about that last inequality? Is it true? Hint: Collect all the terms on the right hand side and think about what you have.
 
  • #27
Office_Shredder
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That looks correct. In the first line there's a + in the right hand side where it's supposed to be a multiplication, but you use it as a multiplication in the second line, so it doesn't cause any problems with the final line.

Now comes trying to prove the inequality that you have left... any ideas on what to do?
 
  • #28
ibaforsale
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what terms are there to collect? it seems to me that they are different, the first las a1b2 and the second has a2b1
 
  • #29
LCKurtz
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Yes, they are all different. I just meant get all the variables on the right side.
 
  • #30
ibaforsale
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oh okay, so i get

2 <= (a12b22 + a22b12) / a1b1a2b2

or if i move everything i get

0 <= a12b22 + a22b12 - 2a1b1a2b2
 
  • #31
LCKurtz
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oh okay, so i get

2 <= (a12b22 + a22b12) / a1b1a2b2

or if i move everything i get

0 <= a12b22 + a22b12 - 2a1b1a2b2

OK. Only the second expression has everything on the right side. So now you have to figure out why that is ##\ge 0##.
 
  • #32
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oh okay, so i get

2 <= (a12b22 + a22b12) / a1b1a2b2
The above won't do you any good.
or if i move everything i get

0 <= a12b22 + a22b12 - 2a1b1a2b2

Or more suggestively, 0 <= a12b22 - 2a1b1a2b2 + a22b12

There's something you need to do with this...
 
  • #33
ibaforsale
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OK. Only the second expression has everything on the right side. So now you have to figure out why that is ##\ge 0##.

is it greater than 0 because the squared terms will always be greater?
 
  • #34
LCKurtz
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is it greater than 0 because the squared terms will always be greater?

How do you know that? Are you just guessing?
 
  • #35
ibaforsale
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my thinking is that squaring something will always be bigger than multiplying by 2
 

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