Is the Dot Product Simply the Square of a Vector's Magnitude?

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Discussion Overview

The discussion centers around the relationship between the dot product of a vector and its magnitude, specifically whether the dot product can be equated to the square of a vector's magnitude. Participants explore the algebraic properties of vectors and the implications of the inner product.

Discussion Character

  • Technical explanation, Conceptual clarification

Main Points Raised

  • One participant asserts that squaring the magnitude of a vector yields the dot product, stating ||v||^2 = v . v.
  • Another participant confirms the algebraic relationship, indicating that ||v|| = sqrt(v . v) is valid.
  • A participant expresses uncertainty about the potential existence of a cosine rule in this context.
  • One participant provides a specific example using vector components, illustrating that the dot product of a vector with itself equals the sum of the squares of its components.
  • Another participant agrees that if the inner product is valid, then the previous statements hold true.
  • A later reply suggests that the cosine rule is not applicable here due to the angle between the vectors being zero.

Areas of Agreement / Disagreement

Participants generally agree on the algebraic relationships involving the dot product and the magnitude of vectors, but there is some uncertainty regarding the relevance of the cosine rule.

Contextual Notes

Some assumptions about the validity of the inner product and the conditions under which these relationships hold are not explicitly stated.

cytochrome
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If you square the magnitude of a vector you get the dot product, correct?

||v||^2 = v . v

Can you also say that

||v|| = sqrt(v . v)?
 
Last edited:
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Of course, basic algebra.
 
Vorde said:
Of course, basic algebra.

Thanks. I didn't know if some weird cosine rule existed in there
 
Okay. Just to cement this:

If ##\vec{v} = <a,b>## and ##\vec{w} = <c,d>## then ##\vec{v} \cdot \vec{w} = ac+bd## and ##\vec{v} \cdot \vec{v} = a^2+b^2##

So if ##|| \vec{v} || ^2 = \vec{v} \cdot \vec{v} = a^2+b^2## then ##\sqrt{|| \vec{v} || ^2} = || \vec{v} || = \sqrt{a^2+b^2}##
 
Hey cytochrome.

If the inner product is valid then all of your statements are true.
 
cytochrome said:
Thanks. I didn't know if some weird cosine rule existed in there

cosine rule cannot bother you here because the angle between "vectors" is zero.
 

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