Is equation of a hyperplane fixed? If I have a $n$ dimensional vector space ( real components ) then a hyperplane will be $n-1$ dimensional. The equation of a hyperplane is defined as $\vec{n}.\vec{x}=\vec{n}.\vec{x_0}$ ( if I am not wrong ) here $\vec{x_0}$ is a position vector on the hyperplane and $\vec{n}$ is the normal unit vector to the hyperplane ( $\vec{x}$ is the variable position vector ). So as   $\vec{n}$ is fixed for a hyperplane I am confused as the hyperplane can have multiple equations since I can change $x_0$ to be some other fixed point on the hyperplane. I know I am doing something stupid but can't figure out.
 A: In $n$-dimensional space $a_1x_1+a_2x_2+\cdots+a_nx_n=0$ (for fixed coefficcients $a_j$'s) define a hyperplane: thinking of the LHS of this eq. as dot product between the fixed point $(a_1,a_2,\ldots,a_n)$ and a variable point $(x_1,x_2,\ldots, x_n)$ this is precisely the set of points perpendicular to that fixed point. When $a_j$'a are uniformly multiplied by a fixed constant the new equation still defines the same locus, so the same hyperplane. We can normalize and write the equation  and so there is only one locus.
A: A hyperplane $H\subseteq V$ is determined by its points and the set of points in a hyperplane $H$ is $H=\lbrace v\in V: \vec{n}\cdot \vec{v}=\vec{n}\cdot\vec{x_0}\rbrace$. If we take two different position vectors $\vec{x_0},\vec{x_1}\in H$, then $\vec{d}=\vec{x_1}-\vec{x_0}$ is parallel to $H$ and hence $\vec{n}\cdot\vec{d}=0$. Then we have 
$$\vec{n}\cdot \vec{v}=\vec{n}\cdot\vec{x_0}=\vec{n}\cdot\left(\vec{x_0}+\vec{d}\right)=\vec{n}\cdot\left(\vec{x_0}+\vec{x_1}-\vec{x_0}\right)=\vec{n}\cdot\vec{x_1}$$ 
We get $H=\lbrace v\in V: \vec{n}\cdot \vec{v}=\vec{n}\cdot\vec{x_1}\rbrace$, so the hyperplane stays the same regardless of the position vector and the equation is not unique; additionally you can scale $\vec{n}$ by any nonzero scalar and still get the same hyperplane. By normalizing of the normal vector (i.e. scaling it to length $1$) and the position vector (for example taking it to be the one with minimal distance to the origin), you can get a standard representation of an hyperplane.
