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

AI Thread Summary
The dot product of a vector is indeed equal to the square of its magnitude, expressed mathematically as ||v||^2 = v · v. This relationship allows for the conclusion that ||v|| can be calculated as the square root of the dot product, ||v|| = sqrt(v · v). For example, if vector v is represented as <a, b>, then its dot product with itself results in a^2 + b^2. The discussion clarifies that there is no need for a cosine rule in this context, as the angle between the vectors is zero. Overall, the mathematical principles governing the dot product and vector magnitudes are confirmed.
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)?
 
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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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