# Thread: A system of cross and dot products

1. ## A system of cross and dot products

This was found in a textbook on classical mechanics so I thought it best to post it here rather then in the calc forum.

$\displaystyle \bold{b}\cdot\bold{v}=\lambda$
$\displaystyle \bold{b}\times\bold{v}=\bold{c}$

$\displaystyle \bold{b}$,$\displaystyle \bold{c}$,and $\displaystyle \lambda$ are known. Express $\displaystyle \bold{v}$ in their terms.

I can't think of which rules would give me an answer. Any help would be appreciated.

2. Sorry!!! Fool me!

Btw: two wonderful quotes.

3. Are they maybe vectors? Then it'll be different to what Also sprach Zarathustra posted.

4. Actually, this does belong to the calculus forum.
Recall that $\displaystyle u\times (v\times w)=(u\cdot w)v-(u\cdot v)w$ so
$\displaystyle b\times (b\times v)=(b\cdot v)b-(b\cdot b)v=b\times c$
Now solve for $\displaystyle v$.

5. Thank you Thus Spake Zarathustra

And thank you Plato I was never aware of that formula. I'll be sure to prove it before I move on.

So the identity you've given when combined with the equations gives

$\displaystyle \lambda \bold{b}-b^2\bold{v}=\bold{b}\times \bold{c}$
which rearranged gives

$\displaystyle \bold{v}=\dfrac{\lambda \bold{b}-\bold{b}\times \bold{c}}{b^2}$

Which matches the answer provided by the text.

6. One small detail. It should be $\displaystyle ||b||^2=b\cdot b$.

7. Oh yes of course. I had figured that notation was standard but I meant $\displaystyle ||\bold{b}||=b$. I'll make it more clear in the future. Thank you.