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Homework Help: Angle Between Two Vectors

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data

    http://www.flickr.com/photos/coolamasta/4563408911/ [Broken]


    2. Relevant equations

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

    3. 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 dunno what to do. please help. tq.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 29, 2010 #2

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    A login is needed to see this page.

    Please post the question here, without pictures.
     
    Last edited by a moderator: May 4, 2017
  4. Apr 29, 2010 #3
    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.
     
    Last edited by a moderator: Apr 29, 2010
  5. Apr 29, 2010 #4
    http://1.bp.blogspot.com/_2w6xphCWyPE/S9ntDt8niJI/AAAAAAAAAE8/6KISjbB_YCU/s1600/soalan1.jpg [Broken]

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

    my lecturer gave me this assignments 3 days ago. i spent 2 days already tried to solving it but still cant 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.
     
    Last edited by a moderator: May 4, 2017
  6. Apr 29, 2010 #5

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    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.
     
  7. Apr 29, 2010 #6
    Sorry for that. This is my work so far.

    http://2.bp.blogspot.com/_2w6xphCWyPE/S9n0ILgCiXI/AAAAAAAAAFU/W1NtWNirRBc/s1600/IMG00296-20100430-0459.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. Apr 29, 2010 #7

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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.
     
  9. Apr 29, 2010 #8

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    If [itex]\theta = \arccos(4/21)[/itex], what is [itex]\cos \theta[/itex]?
     
  10. Apr 29, 2010 #9
    Is it there where i went wrong?
     
  11. Apr 29, 2010 #10

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    Never mind, I see that you get that.

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

    [tex]4=-12 + 3\lambda +8[/tex]

    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.
     
  12. Apr 29, 2010 #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?
     
  13. Apr 29, 2010 #12

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    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.
     
  14. Apr 29, 2010 #13
    I think i get it now. I'll try it again. Thank you.
     
  15. Apr 29, 2010 #14

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    Good luck! Let us know if you run into problems.
     
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