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I Define dot product

  1. Jul 31, 2016 #1
    I have seen a proof for the formula of A.B =
    ||A|| ||B|| cos(theta)[ proof using the diagram and cosine rule]. In the proof they have assumed that distributive property of dot product is right. diagram is given below
    100px-Dot_product_cosine_rule.svg.png c.c =(a-b).(a-b) = a^2 +b^2 -2(a.b) [ here they used distributive law]
    • I have seen another proof for the distributive property of dot product. There they have assumed that A.B = ||A|| ||B|| cos(theta),And used projections. They have used the diagram as given below.
    1280px-Dot_product_distributive_law.svg.png
    And projection of vector B on A is ||B||cos(theta) = B.a^ ( a^ is a unit vector in the direction of a vector)[ this is possible if the formula of dot product is assumed to be right.
    • How they can do this , for proving dot product A.B = ||A|| ||B|| cos(theta) they have assumed distributive property to be right and for prooving distributive property they have assumed dot product to be right.
    Therefore I think that there will be a definition for dot product wether it is A.B = ||A|| ||B|| cos(theta) or A.B = a1a2 +b1b2 +c1c2 (component form). If its a definition then how they have defined it like this.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jul 31, 2016 #2

    blue_leaf77

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    The distributive property of inner product follows immediately from the basic definition of the standard inner product. Given two vectors ##\mathbf v## and ##\mathbf w##, their inner product is ##(\mathbf v, \mathbf w) = \overline v_1 w_1 + \overline v_2 w_2 + \ldots + \overline v_N w_N##. From this, it should be straight forward to see that ##(\mathbf v + \mathbf w,\mathbf z) = (\mathbf v,\mathbf z)+(\mathbf w,\mathbf z)##.
     
  4. Jul 31, 2016 #3

    Svein

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  5. Jul 31, 2016 #4

    robphy

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  6. Aug 1, 2016 #5

    chiro

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    Hey parshyaa.

    The cosine rule is done for the general proof and one uses the results for length in arbitrary R^n.
     
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