Define Dot Product: A.B = ||A|| ||B|| cos(theta)

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SUMMARY

The discussion centers on the definition and proof of the dot product, represented as A.B = ||A|| ||B|| cos(theta). Two proofs are presented: one utilizing the cosine rule and the other employing projections, both of which assume the validity of the other property being proven. The conversation highlights the circular reasoning in proving the distributive property of the dot product while assuming the dot product's definition. The conclusion emphasizes the need for a clear definition of the dot product, whether in terms of the cosine formula or the component form A.B = a1a2 + b1b2 + c1c2.

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parshyaa
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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.
 
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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)##.
 
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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