## Dot product arithmetic

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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 Recognitions: Gold Member Of course, basic algebra.

 Quote by Vorde Of course, basic algebra.
Thanks. I didn't know if some weird cosine rule existed in there

Recognitions:
Gold Member

## Dot product arithmetic

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.

 Quote by cytochrome 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.