# Let a and b denote 2 2d vectors

1. Oct 10, 2013

### ibaforsale

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. Oct 10, 2013

### LCKurtz

Are you asking for our permission?

3. Oct 10, 2013

### ibaforsale

no, a push in the right direction.

4. Oct 10, 2013

### LCKurtz

Well, why don't you be brave and try your own suggestion?

5. Oct 10, 2013

### ibaforsale

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. Oct 10, 2013

### LCKurtz

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. Oct 10, 2013

### ibaforsale

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
8. Oct 10, 2013

### ibaforsale

since theres more terms on the right side can it be said that its greater than the left side?

9. Oct 10, 2013

### LCKurtz

No. The first thing you need to do is use the correct formulas for $\|a\|$ and $\|b\|$.

10. Oct 10, 2013

### Office_Shredder

Staff Emeritus
||a|| is not equal to (a1+a2).

11. Oct 10, 2013

### ibaforsale

isnt ||a|| = Sqrt(a12 + a22)?

12. Oct 10, 2013

### LCKurtz

Yes, but that is not equal to $a_1+a_2$.

13. Oct 10, 2013

### ibaforsale

if theyre both squared cant you just take the sqrt and end up with a1+a2?

14. Oct 10, 2013

### LCKurtz

Does $\sqrt{3^2+4^2} = 3+4$?

15. Oct 10, 2013

### ibaforsale

ahh i see

so then i get

a1b1 + a2b2 ≤ √(a12+a22) + √(b12+b22)

Last edited: Oct 10, 2013
16. Oct 10, 2013

### LCKurtz

Where did that + come from? Fix that and then respond to post #6.

17. Oct 10, 2013

### ibaforsale

my mistake

a1b1 + a2b2 ≤ √(a12+a22) √(b12+b22)

how can i determine the truth of the inequality?

Last edited: Oct 10, 2013
18. Oct 10, 2013

### LCKurtz

Work on it. How might you get rid of the square roots?

19. Oct 10, 2013

### ibaforsale

(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
20. Oct 10, 2013

### LCKurtz

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.