General Equation of an (n-1) Plane from Vector Equation The vector equation of a plane in three dimensions is equal to the expression
\begin{align*}
\mathbf{x}=\mathbf{x}_0+t_1\mathbf{v}_1+t_2\mathbf{v}_2,
\end{align*}
where the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ are not colinear. The general equation of a plane is
\begin{align*}
ax+by+cz=d,
\end{align*}
where $a$, $b$, and $c$ are not all zero. Using the cross product to find the normal vector, and then using point normal form,
\begin{align*}
(\mathbf{x}-\mathbf{x}_0)\cdot(\mathbf{v}_1\times\mathbf{v}_2)=0,
\end{align*}
Therefore, after applying the distributive property of the dot product,
\begin{align*}
\mathbf{x}\cdot(\mathbf{v}_1\times\mathbf{v}_2)=\mathbf{x}_0\cdot(\mathbf{v}_1\times\mathbf{v}_2).
\end{align*} My question is, how do you determine the general equation of an (n-1) plane given the vector equation in higher dimensions?
 A: Assume $v_1,\dots v_{n-1}$ are linearly independent, let $W:=\mathrm{span}(v_1,\dots, v_n)$, and consider the affine hyperplane $x_0+W$.
Now, $x\in x_0+W$ is equivalent to $x-x_0\in W$, i.e. it's linearly dependent on $v_i$'s, which can be expressed by the equation
$$\det(x-x_0,v_1,\dots, v_{n-1})=0$$
or, equivalently,
$$\det(x,v_1,\dots, v_{n-1})=\det(x_0,v_1,\dots, v_{n-1})\,. $$
A: 
The (set of) normal vector(s) to a hyperplane is given by the null space of $V^T$, where the columns of $V$ are the spanning vectors of the hyperplane. This answer explains what is meant by the “spanning vectors of a hyperplane”, and why that is the case. 


A hyperplane $H$ in $\mathbb{R}^n$ is defined to be an $n-1$ dimensional subspace of $\mathbb{R}^n$. This definition may seem against the usual notions of what it means to be a plane, since it requires that $\mathbf{0}$ is an element of $H$. That is, that the plane must go though the origin. But through a simple translation, we can extend this definition.

From the definition we find that $H$ must be spanned by $n-1$ linearly independent vectors $\mathbf{V}_1, \mathbf{V}_2,...\mathbf{V}_{n-1}$ in $\mathbb{R}^n$, we call these “the spanning vectors of the hyperplane”  . So the vector equation describing $H$ is:
$$\mathbf{X}=V\mathbf{T}$$
Where $V$ is the $n$ by $n-1$ matrix whose columns are $\mathbf{V}_1,...,\mathbf{V}_{n-1}$. Moreover, we define that $T=\begin{pmatrix} t_1 \\ ....\\ t_{n-1} \end{pmatrix}$. Let $\mathbf{N} \in \mathbb{R}^n$ be a vector such that:
$$V^{T}\mathbf{N}=\mathbf{0}$$
Or equivalently $\mathbf{N}^{T}V=\mathbf{0}$ since $(AB)^T=B^TA^T$. Then from our parametric equation it easily follows that $\mathbf{N}^{T} \mathbf{X}=\mathbf{N} \cdot \mathbf{X}=0$ by multiplying both sides of our parametric vector equation by $\mathbf{N}^T$. Precisely the equation you are looking for. To calculate an $\mathbf{N}$ that works it’s simply enough to calculate the null space of $V^{T}$ since it was defined that $V^T \mathbf{N}=\mathbf{0}$. 

If instead we had the “translated hyperplane” given by:
$$\mathbf{X}=\mathbf{X_0}+V\mathbf{T}$$
Multiplying by $N^T$ this time gives,
$$\mathbf{N}^T \mathbf{X}=\mathbf{N} \cdot \mathbf{X}=\mathbf{N}^T \mathbf{X_0}=\mathbf{N} \cdot \mathbf{X_0}$$
Or,
$$\mathbf{N} \cdot \mathbf{X}=\mathbf{N} \cdot \mathbf{X_0}$$
Precisely the equation you want. 
