
#1
Jan1413, 06:39 PM

P: 161

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)? 



#3
Jan1413, 07:09 PM

P: 161





#4
Jan1413, 08:07 PM

P: 784

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}## 



#5
Jan1413, 10:11 PM

P: 4,570

Hey cytochrome.
If the inner product is valid then all of your statements are true. 



#6
Jan1513, 04:23 AM

P: 197




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