What is the Angle Between Two Vectors?

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Homework Statement



http://www.flickr.com/photos/coolamasta/4563408911/


Homework Equations



v . w = ||v|| ||w|| cos q

The Attempt at a Solution



i calculate until i got the lambda value is 8/3. then i compare it with 21 and the answer is wrong. i don't know what to do. please help. tq.
 
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coolamasta said:
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Please post the question here, without pictures.
 
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The angle between two vectors v1 = (6,3,-2) and v2 = (-2,lambda,-4) is arccos (4/21). Find the value(s) of lambda. i hope u can understand the questions. thanks in advance.
 
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http://1.bp.blogspot.com/_2w6xphCWyPE/S9ntDt8niJI/AAAAAAAAAE8/6KISjbB_YCU/s1600/soalan1.jpg

http://4.bp.blogspot.com/_2w6xphCWyPE/S9ntPNm08TI/AAAAAAAAAFE/8w4Fsfu8ISA/s1600/soalan3.jpg

my lecturer gave me this assignments 3 days ago. i spent 2 days already tried to solving it but still can't solve it. i manage to solve 3 out of 5 of his questions but the other two i cant. i went to see him this morning n he's not giving me any tips or solution at all.. i tried to find example questions in his tutorial books but did not find anywhere near this questions. please help me. thanks in advance.
 
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coolamasta said:
i calculate until i got the lambda value is 8/3. then i compare it with 21 and the answer is wrong. i don't know what to do. please help. tq.
FIrst off, please do learn to use proper English. Sentences start with a capital letter. So does "I".'tq' is not a word. Stop with the silly txtspeak.

OK, had to get that off my chest.

Now as far as your calculated lambda value, show some work. It's a bit hard to figure out where you went wrong when you didn't show us what you did.
 
Sorry for that. This is my work so far.

http://2.bp.blogspot.com/_2w6xphCWyPE/S9n0ILgCiXI/AAAAAAAAAFU/W1NtWNirRBc/s1600/IMG00296-20100430-0459.jpg
 
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Correction post first:

The problem statement has v1=(6,3,-2). You stated v1=(6,3,2) in post #3. I edited post #3 to reflect the statement.
 
If \theta = \arccos(4/21), what is \cos \theta?
 
Is it there where i went wrong?
 
  • #10
Never mind, I see that you get that.

What you did wrong was to assume that your work meant that

4=-12 + 3\lambda +8

and the same for the denominator.

Look at it this way: 3/6=1/2. certainly neither of 3=1 nor 6=2 is true. Now let's put a variable in there: x/6=1/2. What is x? The solution is found by multiplying both sides by 6, yielding x=6/2=3.

In this problem you have a radical, so at some point it might help to square both sides.

A problem can arise when you do this. Just because you have a solution to the squared expression does not mean you have a solution to the original expression. For example, -1=1 is obviously false, but squaring both sides yields 1=1, obviously true. You will need to double-check your solutions.
 
  • #11
This is the only solution that i know. I've asked everyone around me and they told me they also couldn't find the answer for it. What should I do now?
 
  • #12
What you did was wrong back in elementary school days when you had to learn fractions.

What is the first step often used to solve a problem involving fractions? You want to change the problem to one not involving fractions.
 
  • #13
I think i get it now. I'll try it again. Thank you.
 
  • #14
Good luck! Let us know if you run into problems.
 
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