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

[cytochrome]

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)?

 

[Vorde]

Of course, basic algebra.

 

[cytochrome]

Of course, basic algebra.

Thanks. I didn't know if some weird cosine rule existed in there

 

[Vorde]

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 $\sqrt{|| \vec{v} || ^2} = || \vec{v} || = \sqrt{a^2+b^2}$

 

[chiro]

Hey cytochrome.

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

 

[xAxis]

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.