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Vector help please

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Three units vectors a, b, and c have property that the angle between any two is a fixed angle [tex]\theta[/tex]

    (i) find in terms of [tex]\theta[/tex] the length of the vector v = a + b + c
    (ii) find the largest possible value of [tex]\theta[/tex]
    (iii) find the cosine of the angle [tex]\beta[/tex] between a and v

    2. Relevant equations
    unit vector = vector with length 1unit

    magnitude of vector = [tex]\sqrt{x^2+y^2+z^2}[/tex]

    [tex]\cos \theta = \frac{r_1\cdot r_2}{|r_1||r_2|}[/tex]

    3. The attempt at a solution
    (i) I think I get it right. The answer is [tex]\sqrt{3+6\cos \theta}[/tex]

    (ii) I don't know how to do this. I think [tex]\theta < 90^o[/tex] , but I can't find the exact value

    [tex]\cos \beta = \frac{a\cdot v}{|a||v|}[/tex]

    After some calculation,

    [tex]\cos \beta = \frac{2+\cos \theta}{\sqrt{3+6\cos \theta}}[/tex]

    Can it be simplified further?

    Thanks a lot
  2. jcsd
  3. Oct 12, 2009 #2


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    Re: vector

    i) looks ok

    ii) think about the case when they are all in the same plane...

    iii) shouldn't this be 1 + 2cos(theta) on the numerator?
  4. Oct 12, 2009 #3


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    Re: vector

    Look at (i), and ask yourself for what values of theta can that length even exist? You know that v=a+b+c must be an actual vector, which means it must have an actual length
  5. Oct 12, 2009 #4


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    Re: vector

    though v can be the zero vector, with zero length
  6. Oct 12, 2009 #5
    Re: vector

    Hi lanedance and Office_Shredder

    Ah yes or (iii) it should be 1 + 2cos(theta). I found it but dunno why I wrote 2 + cos(theta) here....:redface:

    For (ii) , The length of v can exist if :

    [tex]3+6 \cos \theta \geq 0[/tex]

    I found the value for [tex]\theta[/tex] = [0o, 120o] U [240o, 360o] for [tex]0^o\leq \theta \leq 360^o[/tex]

    How to continue :confused:

  7. Oct 12, 2009 #6


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    Re: vector

    so you're pretty much there,

    first though, the way to visualise this is to consider all the vectors pointing in the same direction, theta = 0. this is where |v| = 3

    as the angle is increased, imagine the vectors spreading something like a flower opening, keeping the same angle between each, with |v| decreasing. The maximum angle occurs when they are all in a plane, theta = 120, and |v| = 0. Agreeing with the first range of your solution.

    I also think you only need to consider upto 120 (solutions for 120<theta<= 180 do not exist, and above 180 you can just measure the angle the other way)
    Last edited: Oct 12, 2009
  8. Oct 12, 2009 #7
    Re: vector

    Hi lanedance

    Ahh I get it now

    Thanks a lot for you both !!:smile:
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