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Finding the unknown component of a vector that makes a particular angle with a known

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Let u = (1,k) and v = (3,4). Find k such that the angle between u and v is [itex]{\pi}/3[/itex] radians.

    2. Relevant equations

    [itex]u{\bullet}v=||u|| ||v|| cos{\theta}=x_{1}x_{2}+y_{1}y_{2}[/itex]

    [itex]||u||=\sqrt{x^2 + y^2}[/itex]

    3. The attempt at a solution

    Firstly I calculate the length of v and find an expression for the length of u:

    [itex]||u||=\sqrt{1 + k^2}[/itex]

    [itex]||v||=\sqrt{3^2 + 4^2}[/itex]
    [itex]||v||=5[/itex]

    Then I find an expression for the dot product:

    [itex]u{\bullet}v=3+4k[/itex]

    I plug my expressions for the dot product and lengths into the definition of the dot product, and set [itex]\theta[/itex] to [itex]\pi/3[/itex], giving me:

    [itex]3+4k=5cos{(\pi/3)}\sqrt{1+k^2}[/itex]

    as [itex]cos{(\pi/3)}=1/2[/itex], I can substitute [itex]cos{(\pi/3)}[/itex] in my equation for 1/2, giving:

    [itex]3+4k=5/2\sqrt{1+k^2}[/itex]

    I rearrange and expand brackets to get:

    [itex]6+8k=5\sqrt{1+k^2}[/itex]

    I then square both sides to get rid of the square root, then expand brackets:

    [itex](6+8k)^2=25+25k^2[/itex]
    [itex]36 + 96k + 64k^2=25+25k^2[/itex]

    I move everything to one side:

    [itex]39k^2+96k+11=0[/itex]

    Using the quadratic formula, I get the answer that k= -2.341058209 or -0.1204802515. Clearly one of these is wrong (or both), as there can't be two angles in the same quadrant that make an angle of [itex]\pi/3[/itex] with a vector. But both of these values do satisfy the equation for the dot product of these two angles when the angle between them is [itex]\pi/3[/itex]. Have I done something wrong?

    PS: I wasn't sure if this was the right forum, as both maths forums seem to be calculus-orientated. Sorry if it's not in the right place!
     
  2. jcsd
  3. Apr 11, 2012 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    hi phosgene! :smile:
    no, you did everything fine :smile:

    however, when you squared both sides (which was correct), you automatically introduced an extra solution (for cos = -1/2, ie 2π/3), and you now need to check which of your two solutions is for π/3 ! :wink:

    (ie just check that the dot-product is positive)
     
  4. Apr 11, 2012 #3
    Re: Finding the unknown component of a vector that makes a particular angle with a kn

    Thanks:) I just found out using google that the dot product is negative if the angle between the vectors is greater than 90 degrees. I had no idea that this was the case (it wasn't mentioned in the lectures). So it's crystal clear now. Thanks again!
     
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