Hyperplane conversion between vector form and cartesian form Vector form to Cartesian form
In $\mathbb{R^2}$ a $\mathbb{R^1}$ object can be expressed as
$$\text{vector form: }s\begin{pmatrix}a\\b\end{pmatrix}+\begin{pmatrix}c\\d\end{pmatrix}$$
$$\text{cartesian form: }\begin{vmatrix}a&b\\x-c&y-d\end{vmatrix}=0$$
In $\mathbb{R^3}$ a $\mathbb{R^2}$ object can be expressed as
$$\text{vector form: }s\begin{pmatrix}a\\b\\c\end{pmatrix}+t\begin{pmatrix}d\\e\\f\end{pmatrix}+\begin{pmatrix}g\\h\\i\end{pmatrix}$$
$$\text{cartesian form: } \begin{vmatrix}a&b&c\\d&e&f\\x-g&y-h&z-i\end{vmatrix}=0$$
$\vdots$

In $\mathbb{R^n}$ a $\mathbb{R^{n-1}}$ object can be expressed as
$$\text{vector form: }s_1\begin{pmatrix}a_1\\a_2\\ \vdots\\a_n\end{pmatrix}+s_2\begin{pmatrix}b_1\\b_2\\ \vdots\\b_n\end{pmatrix}+\dots+s_{n-1}\begin{pmatrix}c_1\\c_2\\ \vdots\\c_n\end{pmatrix}+\begin{pmatrix}d_1\\d_2\\ \vdots\\d_n\end{pmatrix}$$
$$\text{cartesian form: } \begin{vmatrix}a_1&a_2&\dots&a_n\\b_1&b_2&\dots&b_n\\\vdots&&&\vdots\\c_1&c_2&\dots&c_{n}\\x_1-d_1&x_2-d_2&\dots&x_n-d_n\end{vmatrix}=0$$

I always use this fomular to do the conversion, but is it true for all $n\in[2,\infty)\cap\mathbb{N}$.
What I tried is using induction, for Base case, It's not hard to show it hold when n=2, but i don't know how to write the inductive step, or is this easier to do with direct prove?
Any help or hint or suggestion would be appreciated.

Cartesian form to Vector form
If given cartesian form in $\mathbb{R^2}$:
$$ax+by+c=0$$
when $c\neq0$, $X,Y$ are intersections of the line and aixs:
$$X=\begin{pmatrix}-\frac{c}{a}\\0\end{pmatrix} \text{ and } Y=\begin{pmatrix}0\\-\frac{c}{b}\end{pmatrix}$$
Therefore the line can be expressed as:
$$\text{vector form: }s(X-Y)+X$$
Since $X=Y$ when $c=0$, the method above no longer works.
But $N_1=\begin{pmatrix}-\frac{1}{a}\\\frac{1}{b}\end{pmatrix}, N_2=\begin{pmatrix}\frac{1}{a}\\-\frac{1}{b}\end{pmatrix}$ are normal to the normal of the line.
Choose any 1 as direction vector have:
$$\text{vector form: }sN_1$$
If given cartesian form in $\mathbb{R^3}$:
$$ax+by+cz+d=0$$
when $d\neq0$, $X,Y,Z$ are intersections of the plane and aixs:
$$X=\begin{pmatrix}-\frac{d}{a}\\0\\0\end{pmatrix}, Y=\begin{pmatrix}0\\-\frac{d}{b}\\0\end{pmatrix},Z=\begin{pmatrix}0\\0\\-\frac{d}{c}\end{pmatrix}$$
Therefore the plane can be expressed as:
$$\text{vector form: }s(X-Y)+t(X-Z)+X$$
Since $X=Y=Z$ when $d=0$, the method above no longer works.
But $N_1=\begin{pmatrix}-\frac{2}{a}\\\frac{1}{b}\\\frac{1}{c}\end{pmatrix}, N_2=\begin{pmatrix}\frac{1}{a}\\-\frac{2}{b}\\\frac{1}{c}\end{pmatrix}, N_3=\begin{pmatrix}\frac{1}{a}\\\frac{1}{b}\\-\frac{2}{c}\end{pmatrix}$ are normal to the normal of the plane.
Choose any 2 as direction vectors have:
$$\text{vector form: }sN_1+tN_2$$
$\vdots$

More generally, If given cartesian form in $\mathbb{R^n}$:
$$a_1x_1+\dots+a_nx_n+b=0$$
Have intersections on $x_1$ that $X_1=\begin{pmatrix}-\frac{b}{a_1}\\0\\\vdots\\0\end{pmatrix}$
And $N_1=\begin{pmatrix}-\frac{n-1}{a_1}\\\frac{1}{a_2}\\\vdots\\\frac{1}{a_n}\end{pmatrix}, N_2=\begin{pmatrix}\frac{1}{a_1}\\-\frac{n-1}{a_2}\\\frac{1}{a_3}\\\vdots\\\frac{1}{a_n}\end{pmatrix},\dots, N_n=\begin{pmatrix}\frac{1}{a_1}\\\frac{1}{a_2}\\\vdots\\\frac{1}{a_{n-1}}\\-\frac{n-1}{a_n}\end{pmatrix}$ are normal to the normal.
Choose any $n-1$ as direction vectors have: $$\text{vector form: }s_1N_1+\dots+s_{n-1}N_{n-1}+X_1$$


Note that the vector form and the cartesian form are equivalent only if the $n-1$ vectors $(a_1\;\ldots\;a_n)^T \;\ldots\;(c_1\;\ldots\;c_n)^T$ are linearly independent. Otherwise, the vector form would produce something of degree less than $n-1,$ while the cartesian form would produce the complete $\mathbb{R}^n.$ So we assume the linear independence of $(a_1\;\ldots\;a_n)^T \;\ldots\;(c_1\;\ldots\;c_n)^T$ in the following.
If the cartesian form is given, we know that the $n$ vectors $(a_1\;\ldots\;a_n)^T \;\ldots\;(c_1\;\ldots\;c_n)^T,\;(x_1-d_1\;\ldots\;x_n-d_n)^T$ are linearly dependent, which means that there are $n$ real numbers $t_1,\ldots, t_n$, at least one of them $\neq 0,$ such that
$$
t_1 \begin{pmatrix} a_1 \\ \vdots \\ a_n\end{pmatrix}
+t_2 \begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}
+\ldots
+t_{n-1} \begin{pmatrix} c_1 \\ \vdots \\ c_n\end{pmatrix}
+t_{n} \begin{pmatrix} x_1-d_1 \\ \vdots \\ x_n-d_n\end{pmatrix}
=0
$$
Assume $t_n=0.$ Then one of the numbers $t_1,\ldots, t_{n-1}$ must be $\neq 0,$ because we assumed the $t_i\neq 0$ for at least one $i\in \{1,\ldots,n\}.$ This would mean that we had the following:
$$
t_1 \begin{pmatrix} a_1 \\ \vdots \\ a_n\end{pmatrix}
+t_2 \begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}
+\ldots
+t_{n-1} \begin{pmatrix} c_1 \\ \vdots \\ c_n\end{pmatrix}
=0
$$
But this is a contradiction to the linear independence of $(a_1\;\ldots\;a_n)^T \;\ldots\;(c_1\;\ldots\;c_n)^T.$ Therefore, $t_n\neq 0.$
We can now set $t_i = -t_n s_i$ for $i\in\{1,\ldots, n-1\}$, divide the equation by $-t_n$  and we get
$$
\begin{pmatrix} x_1 \\ \vdots \\ x_n\end{pmatrix}
=
s_1 \begin{pmatrix} a_1 \\ \vdots \\ a_n\end{pmatrix}
+s_2 \begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}
+\ldots
+s_{n-1} \begin{pmatrix} c_1 \\ \vdots \\ c_n\end{pmatrix}
+\begin{pmatrix} d_1 \\ \vdots \\ d_n\end{pmatrix}
$$
which shows that the given element of the hyperplane can also be expressed in terms of the vector form.
If the vector form is given, we have
$$
s_1 \begin{pmatrix} a_1 \\ \vdots \\ a_n\end{pmatrix}
+s_2 \begin{pmatrix} b_1 \\ \vdots \\ b_n\end{pmatrix}
+\ldots
+s_{n-1} \begin{pmatrix} c_1 \\ \vdots \\ c_n\end{pmatrix}
=
\begin{pmatrix} x_1-d_1 \\ \vdots \\ x_n-d_n\end{pmatrix}
$$
which shows that the rows of the determinant are linearly dependent, which in turn shows that the value of the determinant is $0.$
