If I have a vector v in $\mathbb{R}^n$, the set of vectors $ \{\vec x\in \mathbb{R}^n\mid \vec {x} \cdot \vec {v}=0 \} $ is a vector space of dimension $n-1.$
I understand it just as a linear equation (restriction) that reduces the dimension, but, is $ \{\vec x\in \mathbb{R}^n\mid \vec {x} \cdot \vec {v}=c\}$ also a vectorspace if $c\neq 0$?
On
$V= \{\vec x\epsilon \mathbb{R}^n| \vec {x} \cdot \vec {v}=c\}$ is not a vector space.
To define a vector space, we need a non empty set and a field and two operations (one allows us to add two vectors and other allows us to scale a vector).
Here, you consider the vector space operation of $\mathbb{R^n}$.
It is easy to verify that it violates the very first property of vector space (namely closed under addition)
$x, y \in V $
Is this implies $x+y \in V$ ?
$ \vec x \cdot \vec v = c$ and $ \vec y \cdot \vec v = c$
$\vec{ (x+y)} \cdot \vec v = \vec x \cdot \vec v + \vec y \cdot \vec v = 2c$
Hence, if $c\neq 0$ then, $x+y \notin V$
The short answer is no.
I am not a linear algebra expert, but there are two simple rules of a vector space.
If $\vec{x} \in V$, then $a\vec{x} \in V$ too, for constant $a$.
We already have an issue, because if $a=0$, then the dot product cannot be nonzero.