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???