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Angle between vectors

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data

    Let C be a straight path from (0,0) to (5,5) and let F=(y-x+2)i + (sin(y-x)+2)j.


    At each point of C, what angle does F make with a tangent vector to C ?

    3. The attempt at a solution

    Well the displacement vector for C going from (0,0) to (5,5) is simply
    C=5i+5j. The tangent vector to this would be parallel to this vector, since C is a straight line. So my tangent vector is also 5i+5j. (Does this make sense?) Now to find the angle between F and C, i chose to do a dot product between them.

    The dot product yields following result.

    F.C = 5y-5x+5sin(y-x).

    Since y=x in the displacement vector C, can i simply say that y=x, so the dot product is zero. Therefore the tangent vector and F vector are perpendicular to each other.

    Thanks in advance.
     
  2. jcsd
  3. Dec 4, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You are using a character that shows only as a square on my internet reader. Did you mean find the angle F itself makes with the tangent to C?

    Yes, since C is a straight line a "tangent" vector is just a vector in the same direction as C: 5i+ 5j or, for that matter, just i+ j. However, the dot product F.C is NOT 5y-5x+5sin(y-x). You have forgotten the "+2" in each component. F. C= 5y- 5x+ 10+ 5sin(x-y)+ 10 which is equal to 20 on C. If the problem is just to find the angle the vector field F(x,y) makes with the line y= x, on y= x, F(x,y)= 2i+ 2j is always in the same direction as C.
     
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