Does every hyperplane have a unique normal vector (up to non-zero scalar multiplication)? I am able to prove that the conjecture below is true in the $\mathbb{R}^3$ case, but I was wondering if the general case is true and if it is, how we might (nicely) prove it.
Conjecture. Suppose $p$ is a hyperplane in $\mathbb{R}^{n}$. Then $p$ has a normal vector and moreover, this normal vector is unique (up to non-zero scalar multiplication).

My definitions:
Definition 1. Given $\mathbf{x}=\left(x_1,x_2,\dots,x_n\right)$ and $\mathbf{y}=\left(y_1,y_2,\dots,y_n\right)$, we define: $$\mathbf{x}\cdot \mathbf{y}=\sum_{i=1}^n x_i y_i.$$
Definition 2. A hyperplane in $\mathbb{R}^{n}$ is any set of points that can be written as $$\left\{ \mathbf{x}\in\mathbb{R}^{n}:\mathbf{a}\cdot\mathbf{x}=c\right\},$$ for some $\mathbf{a}\in\mathbb{R}^{n}\setminus\left\{ \mathbf{0}\right\} $ and some $c\in\mathbb{R}$.
Definition 3. We say that $\mathbf{b}=\left(b_{1},b_{2},\dots,b_{n}\right)\neq\mathbf{0}$ is a normal vector of the plane $p$  if: $$\mathbf{x},\mathbf{y}\in p \implies \mathbf{b}\cdot\left(\mathbf{x}-\mathbf{y}\right)=0.$$
Related.
 A: Short answer, yes.
In fact your "definition 2" gives $\mathbf a$ as the coefficients of the equation of a hyperplane as a nonzero vector, and this nonzero vector is a normal to that hyperplane as called for in "definition 3".
The algebra to prove this is very easy.  Write down the equations, satisfied by $x$ and $y$ of the hyperplane respectively, and take the difference.  We immediately have $\mathbf a \cdot (x-y) = 0$ as the constant $c$ cancels.
The argument goes the other way, to show that when $\mathbf b$ in "definition 3" is given, it also serves as a set of coefficients $\mathbf a$ in "definition 2".  That is, suppose a hyperplane $p$ has a normal vector $b$ as described above, and fix some point $y$ in that hyperplane:
$$ \mathbf b \cdot (x-y) = 0 \; \text{ for any } x\in p$$
Expanding the above gives $\mathbf b \cdot x = \mathbf b \cdot y$, and now that $y$ is fixed we get an equation of the form in "definition 2" where constant $c = \mathbf b \cdot y$:
$$ \mathbf b \cdot x = c $$
Therefore any $\mathbf a$ that defines the hyperplane equation in "definition 2" will serve as the normal vector $\mathbf b$ defined in "definition 3", and any normal vector $\mathbf b$ that satisfies "definition 3" will serve as the vector of coefficients $\mathbf a$ in "definition 2".  As a result these quantities have the same uniqueness property, namely that they are unique only up to taking a nonzero scalar multiple of a vector.

There seems to be a lesson here about rank or dimension of a subspace.  The normal vector to a hyperplane is unique (up to a nonzero scalar multiple) in $\mathbb R^n$ because the hyperplane is an $n-1$ dimensional thing, and this leaves only one dimension for things to be perpendicular to this.
So the uniqueness you want to show depends on making use of the $n-1$ dimensional nature of the hyperplane.  The easiest way to discuss this depends to a great extent on your background in linear algebra.
From the perspective of the one linear equation $\mathbf a \cdot x = c$, we see that the matrix form of this equation has a coefficient matrix of rank one (because $\mathbf a$ is nonzero).  Therefore the solutions to this equation (via the principle of superposition) consist of a particular solution to the inhomogenous equation plus all solutions of the homogenous equation $\mathbf a \cdot x = 0$.  Thus we have an $n-1$ dimensional "affine variety" of solutions.
But let's consider the uniqueness strictly from the perspective of the normal vector.  To simplify things, the normal vectors to the hyperplane $\mathbf a \cdot x = c$ are exactly the same as the normal vectors to the subspace $\mathbf a \cdot x = 0$.  This subspace has a basis of $n-1$ linearly independent vectors.  There is a one dimensional subspace orthogonal to these, (orthogonal is another word for perpendicular or normal), and that subspace is generated by vector $\mathbf a$.
Taken together the basis for the subspace $\mathbf a \cdot x = 0$ and the normal vector $\mathbf a$ give us a basis for all of $\mathbb R^n$.  The latter (because it is generated by one vector) is one dimensional, and the former is $n-1$ dimensional (so that the dimension of $\mathbb R^n$ is $n$ as we know).
This is a special case of the Rank-Nullity Theorem, which says the dimension of the row space of the coefficient matrix of the linear "system" $\mathbf a \cdot x = 0$ plus the dimension of the nullspace (the solutions/points of the hyperplane $\mathbf a \cdot x = 0$) is equal to the dimension $n$ of the domain $\mathbb R^n$.
Now the statement that all the nonzero normal vectors to the hyperplane are related by $\mathbf b = k\mathbf a$ for nonzero scalars $k$ is exactly the proposition that the subspace generated by one of these normal vectors is one dimensional.  In other words Span($\mathbf b$) = Span($\mathbf a$) forces $\mathbf b$ to be a nonzero multiple of $\mathbf a$ and vice versa.
