# Dot product arithmetic

1. Jan 14, 2013

### 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)?

Last edited: Jan 14, 2013
2. Jan 14, 2013

### Vorde

Of course, basic algebra.

3. Jan 14, 2013

### cytochrome

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

4. Jan 14, 2013

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

5. Jan 14, 2013

### chiro

Hey cytochrome.

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

6. Jan 15, 2013

### xAxis

cosine rule cannot bother you here because the angle between "vectors" is zero.