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Linear Algebra Angle Def

  1. Apr 1, 2005 #1
    Hi! I cannot get this definition of angle a:
    cos a = u * v / (||u|| * ||v||)
    and how it match up with
    u * v = ||u|| * ||v|| * cos a
  2. jcsd
  3. Apr 1, 2005 #2


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    What do you mean you don't get it...?

    [tex] \vec{u}\cdot\vec{v}=\left|\left|\vec{u}\right|\right| \cdot \left|\left|\vec{v}\right|\right| \cos\alpha [/tex]

    is the expression of the scalar product of 2 vectors,when you know their modulus & the angle between their directions...

  4. Apr 1, 2005 #3
    From [tex] cos(\alpha) = \frac{u\bullet v}{|u||v|} [/tex]

    To [tex] {cos(\alpha)|u||v| = u\bullet v[/tex]

    Just multiply both sides by [tex] |u||v| [/tex]
  5. Apr 1, 2005 #4


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    I'm sure there's something devious in your post... :uhh:


    EDIT:Not anymore,you edited it... :wink:
  6. Apr 1, 2005 #5

    I switched the wrong quantity, I always do that with latex.
  7. Apr 2, 2005 #6
    Hmm. I formulated me quite wrong. How come, if u = (x1, y1), v = (x2, y2), u * v = x1x2 + y1y2 = ||v|| ||u|| cos a
  8. Apr 2, 2005 #7
  9. Apr 2, 2005 #8
    there are two formulas for scalar product:

    1- *[v] = xx' + yy' where (x,y) and [v](x',y')

    2- *[v] = u*v*cos(,[v])

    they usually give you in exercises the coordinates of the two vectors and ask you to find their angle
  10. Apr 2, 2005 #9
    But, indeed, I've concluded from MathWorld, which I should have checked out right the way of cource, there's not two definition of the dot product. One can from the definition involving cosinus derive the other one, which strangely Schamu's Outline of Linear Algebra state is The definition. Only for complicate the matter in my limit brain, they also define angle as cos a = u * v / (||u|| ||v||). So I'm not as mad you may think I am.
  11. Apr 2, 2005 #10
    Well, if you approach it from the perspective of vector spaces (as you probably are in your linalg course), then you usually define a vector space with an inner product to be a vector space [itex]V/\mathbb{R}[/itex] with an operation [itex]( , ): V\times V \longrightarrow \mathbb{R}[/itex] such that [itex]\forall \ a, b, c \in V[/itex] and [itex]x \in \mathbb{R}[/itex],

    1. [itex](a, b) = (b, a)[/itex]
    2. [itex](b + c, a) = (b, a) + (c, a)[/itex]
    3. [itex](xa, b) = x(a, b)[/itex]
    4. [itex](a, a)[/itex] is definite positive, ie. [itex](a, a)\geq 0[/itex] with equality iff [itex]a = 0[/itex].

    We then define

    [tex] \|a\| = \sqrt{(a, a)}[/tex]

    which we call the magnitude or length or modulus of [itex]a[/itex].

    Now say we choose [itex]a, b[/itex] such that [itex]\|a\| =\|b\| = 1[/itex]. Then

    [tex](a-b, a-b) = (a, a) - 2(a, b) + (b, b) = \|a\|^2 - 2(a, b) + \|b\|^2 \geq 0[/tex]

    with the last inequality by 4., so

    [tex]-2(a, b) \geq -\|a\|^2 - \|b\|^2 = -1^2 - 1^2 = -2 = -2\|a\|\|b\|[/tex]

    [tex] \Longrightarrow (a, b) \leq \|a\|\|b\|[/tex]

    and thus for any [tex]c, d[/itex] we note that

    [tex] \biggr\|\frac{c}{\|c\|}\biggr\| = \biggr\| \frac{d}{\|d\|} \biggr\| = 1[/tex]

    and so by the previous argument,

    [tex] \frac{(c, d)}{\|c\|\|d\|} = \left(\frac{c}{\|c\|}, \frac{d}{\|d\|}\right) \leq \biggr\|\frac{c}{\|c\|}\biggr\|\biggr\|\frac{d}{\|d\|}\biggr\| = \frac{\|c\|\|d\|}{\|c\|\|d\|}[/tex]

    [tex] \Longrightarrow (c, d) \leq \|c\| \|d\| \; \forall c, d \in V[/tex]

    which is the well-known Cauchy-Schwarz inequality.

    From here, we very simply define that

    [tex] (c, d) = \|c\|\|d\| \cos{\theta}[/tex]

    it's just like any other definition you ever will encounter. You can then derive all the properties of the [itex]\cos[/itex] function.
    Last edited: Apr 2, 2005
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