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Calculating a theta using dot product in 3D coordinate

  1. Dec 19, 2013 #1

    Tah

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    xc.jpg

    I'm so confused about finding an angle, theta in this illustration.

    With having three coordinate information, how can I calculate the theta using dot product?

    I would easily find the angle by using trigonometric formula if I ignore the z-axis.

    But I want to solve this problem with 3-dimensional coordinate system by using dot product formula.

    Please help me confused.
     
  2. jcsd
  3. Dec 19, 2013 #2

    ShayanJ

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    Gold Member

    There are two definitions for dot product of two vectors.we have [itex] \vec{u}\cdot\vec{v}=u_xv_x+u_yv_y+u_zv_z=uv\cos{\theta} [/itex].
    I think your answer is clear now!
     
  4. Dec 19, 2013 #3

    Tah

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    Thanks

    Yes, I've already known that formula but could not understand exactly.

    Could you help me more detail about such variables?

    I was wondering about what is u_x, v_x, u_y, v_y.... in this case.

    I think 'u' is (x1-x2, y1-y2, z1-z2) and 'v' is (x2-x3, y2-y3, z2-z3). Is this correct?

    Thank you for your support :)
     
  5. Dec 19, 2013 #4

    ShayanJ

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    [itex] u_x,v_y,... [/itex] x,y and z components of the vectors u and v.They can be anything! [itex] (x_1-x_2, y_1-y_2, z_1-z_2) and (x_2-x_3, y_2-y_3, z_2-z_3) [/itex] are just particular examples when u and v are displacement vectors.You just have to find the components of the vectors and then their magnitudes.
     
  6. Dec 19, 2013 #5

    K^2

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    In this example, yes. And when Shyan writes ##uv\ cos\theta##, here u and v are norms of the two vectors. Just in case it wasn't clear.
     
  7. Dec 19, 2013 #6

    Tah

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    Thanks

    I have one more question

    xc.jpg

    In this case, should I just do 180' minus the theta calculated by using dot product formula to find the theta prime(')?

    Thanks!!
     
  8. Dec 19, 2013 #7

    K^2

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    Yes.

    There exists a plane containing both vectors, and in that plane, all the planar geometry you know applies. So θ and θ' are complimentary angles, and therefore, their sum is 180°.
     
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