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)?
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}##