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Find vectors making a certain angle with given vectors

  1. Aug 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Find all unit vectors in R4 making an angle of [itex]\pi/3[/itex] with the three vectors
    A=(1,1,-1,-1)
    B=(1,-1,1,-1)
    C=(1,-1,-1,1)

    2. Relevant equations
    u.v=|u||v|cos[itex]\Theta[/itex]


    3. The attempt at a solution
    using V=(w,x,y,z) as the vector we are trying to find, I solved the above equation for all three vectors A,B and C obtaining simultaneous equations to solve, however I obtained
    x=y=z and w2-wx-x2=0
    I am not sure how to solve this to find all unit vectors.
    Also is there a method of doing this that I am unaware of?
     
  2. jcsd
  3. Aug 8, 2011 #2

    vela

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    How did you get w2-wx-x2=0? Did you use the fact that V is a unit vector?
     
  4. Aug 9, 2011 #3
    I haven't yet considered the unit vector part, I was just going to divide the answer by the modulus.

    I obtained this results as |u|cos[itex]\Theta[/itex] is 2*1/2=1 for all the vectors A,B&C therefore right hand side of the equation for all the vectors is |v|, so we get v.u=|v| and I took

    |v| = [itex]\sqrt{}w2+x2+y2+z2[/itex]

    I put C.u=B.u to obtain w-x-y+z=w-x+y-z which gave me that y=z

    Then A.u=B.u to obtain w+x-y-z=w-x+y-z which gave me x=y

    Then A.u=C.u to give w+x-y-z=w-x-y+z which gives x=z

    so if you take x, y & z to all equal x (for example)

    and use any of the equations we get w-x=[itex]\sqrt{}w2+3x2[/itex]
    from this I got the w2-wx-x2=0
     
  5. Aug 9, 2011 #4

    vela

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    Recheck your algebra. The second equation doesn't follow from the first.

    Also, if you use the fact that |V|=1, the first equation will give you w=1+x.
     
    Last edited: Aug 9, 2011
  6. Aug 9, 2011 #5

    Okay, sorry I got the algebra wrong, I do not obtain w2-wx-x2=0 instead, I get w=-x. Which would be a solution, however, then fact vela pointed out w=1+x contradicts this, so what value am I supposed to use?
     
  7. Aug 9, 2011 #6

    vela

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    How'd you get w=-x? I got wx+x2=0. Try again!
     
  8. Aug 9, 2011 #7
    Yes, I obtained this too, the you get x(w+x)=0
    Therefore you get either x=0 or w+x=0 which leads to my answer w=-x
     
  9. Aug 9, 2011 #8

    vela

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    Heh. I didn't recognize the root. But there's no problem. You have two equations and two unknowns:
    \begin{align*}
    w+x & = 0 \\
    w-x & = 1
    \end{align*}
    You can solve that.

    Also, don't just ignore the x=0 solution as well.
     
  10. Aug 9, 2011 #9
    therefore we get w=1/2 therefore x=-1/2 y=-1/2 and z=-1/2

    Also using x=0 we get w=0, y=0 & z=0
    so when it asks for all the unit vectors there must only be one, as the second one has a modulus of zero. So are they no more unit vectors apart from
    (0.5,-0.5,-0.5,-0.5)?
     
  11. Aug 9, 2011 #10

    vela

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    You didn't handle the x=0 case correctly. In particular, it doesn't follow that w=0 from x=0.
     
  12. Aug 10, 2011 #11
    Yes, I see it now, if x=0, then the |v|=1 still so we can use the equation w-x=1.
    This implies that w=1, and as x=y=z=0
    We get the second unit vector (1,0,0,0)

    So there are two unit vectors for this answer. Thank you for all your help :)
     
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