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Planar angle between two vectors

  1. Sep 12, 2015 #1
    I have two vectors:
    f1 = (x1,y1,z1)
    f2 = (x2,y2,z2)
    The origins are placed together and they are of the same coordinate system.
    They make a plane between the two vectors and the origin.
    How do you find the angle between the two vector on that plane?
  2. jcsd
  3. Sep 12, 2015 #2


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    [tex]\vec f_1\cdot \vec f_2 = \Vert \vec f_1\Vert \cdot \Vert \vec f_2\Vert \cdot \cos(\alpha)[/tex]
    (i.e., compute the scalar product, divide it by the vector norms, and you get the cosinus of the angle you are searching.)
  4. Sep 12, 2015 #3


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    It so happens I just wrote these notes:

    For vector spaces over R, the dot product lets us measure lengths and angles. I.e. if we think of the axes spanned by the standard basis vectors as mutually perpendicular, or “orthogonal”, then by Pythagoras the square of the length |v| of a vector v = (a1,...,an) is just |v|^2 = a1^2+...+an^2 = v.v. Moreover if v = (a1,...,an), and w = (b1,...,bn) are any two vectors, then |v|,|w|, and |v-w| are the length of the sides of the triangle they determine, and hence by the law of cosines |v-w|^2 = |v|^2 + |w|^2 - 2|v||w|.cos(C), where C is the angle between v and w. Expanding the left side as |v-w|^2 = (v-w).(v-w) = v.v - 2v.w + w.w, we get v.w = |v||w|.cos(C). Since cos(π/2) = 0, this implies that v.w = 0 if (and only if) v and w are perpendicular. This is quite useful.
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