# vectors

• Mar 4th 2014, 02:07 PM
Pipita
vectors
I want to proof that, given vector a non zero and a vector b, so
$\displaystyle a.b=0 <=> \exists c: c\times a=b$

I know how to proof <=

but how to proof =>?
My try:
$\displaystyle a.b=0 => \exists c: c\times a=b$
$\displaystyle <=> \sim \exists c: c\times a=b => \sim a.b=0$
$\displaystyle <=> \forall c: c\times a \neq b => a.b\neq 0$

$\displaystyle Supose \forall c: c\times a \neq b$
I tried to prove $\displaystyle a.b \neq 0$

Now $\displaystyle Supose \forall c: c\times a = b$
so b is orthogonal to both a and c
so $\displaystyle a.b=0$

So if $\displaystyle \forall c: c\times a \neq b$ we have $\displaystyle a.b \neq 0$ my doubt is, my I affirm this???
• Mar 4th 2014, 02:46 PM
Plato
Re: vectors
Quote:

Originally Posted by Pipita
I want to proof that, given vector a non zero and a vector b, so
$\displaystyle a.b=0 <=> \exists c: c\times a=b$
I know how to proof <=
but how to proof =>?

Assume that $a\cdot b=0$ and $a\ne 0$. Now define $c=\dfrac{-(b\times a)}{||a||^2}$

Does that work?
• Mar 4th 2014, 08:06 PM
Informalsports
Re: vectors