# Dot Product

1. Jul 13, 2008

### danago

If $$\vec{u},\vec{v} \in R^n$$, find $$\vec{u}.\vec{v}$$ given that $$|\vec{u}+\vec{v}|=1$$ and that $$|\vec{u}-\vec{v}|=5$$.

When i first looked at this i thought i knew how to do it, but i got a bit stuck. I started by finding the dot product:

$$\begin{array}{l} (\overrightarrow u - \overrightarrow v ) \cdot (\overrightarrow u - \overrightarrow v ) = \left| {\overrightarrow u } \right|^2 + \left| {\overrightarrow v } \right|^2 - 2\overrightarrow u \cdot \overrightarrow v \\ \therefore \left| {\overrightarrow u - \overrightarrow v } \right|^2 = \left| {\overrightarrow u } \right|^2 + \left| {\overrightarrow v } \right|^2 - 2\overrightarrow u \cdot \overrightarrow v \\ \therefore \overrightarrow u \cdot \overrightarrow v = \frac{{\left| {\overrightarrow u } \right|^2 + \left| {\overrightarrow v } \right|^2 - \left| {\overrightarrow u - \overrightarrow v } \right|^2 }}{2} \\ \end{array}$$

And thats where im stuck. Im sure its something simple, but i haven't managed to see it

2. Jul 13, 2008

### HallsofIvy

Staff Emeritus
You are given both $|\vec{u}-\vec{v}|$ and $|\vec{u}+\vec{v}|$ but don't seem to have used the value of $|\vec{u}+ \vec{v}|$. Did you consider looking at $(\vec{u}-\vec{v})\cdot(\vec{u}+\vec{v})= |u|^2- |v|^2$?

3. Jul 13, 2008

### rootX

I couldn't find relationship between |u|^2-|v|^2

(u-v).(u-v) = u^2+v^2-2u.v
(u+v).(u+v) = u^2+v^2+2u.v

now, it's just one step since you know both left side values

4. Jul 15, 2008

### danago

Oh so now i can just solve by eliminating u^2+v^2?

Thanks for the help guys

5. Jul 15, 2008

### rootX

if that answers the question ...