I have the following equivalent representations for a hyperplane:
$H:=\{x\in\mathbb{R}^n\vert w^T\cdot x=d\},\quad w\in\mathbb{R}^n,d\in\mathbb{R}$ fixed
and
$H_2=\left\{x_0+\sum_{i=1}^{n-1}\lambda_i v_i\,\middle\vert\,\lambda_i\in\mathbb{R}\right\}\subseteq\mathbb{R}^n$ with $v_i$ linearly independent and $x_0\in\mathbb{R}^n$ fixed
In words the first representation of a hyperplane is described by an fixed angle represented by $d$ and a fixed vector given by $w$ so that every $x$ with the scalar product $w^T\cdot x=d$ has to lay on the same hyperplane. So all points belonging to the hyperplane are in $H$.
The $x_0$ in $H_2$ should be equivalent to the $w$ in $H$ and the sum represents a point in the hyperplane as combination of $n-1$ linearly independent vectors, since the vectors are fixed all points have to be on the same hyperplane.
I can describe this equivalency easily with words but I am not able to show it mathematically. Could you please help me with this task?
We need that $w\neq 0$, and that $v_1, v_2, \ldots, v_{n-1}$ are linearly independent and all orthogonal to $w$ (implying among other things that $\{w, v_1, v_2, \ldots, v_{n-1}\}$ is a basis of $\Bbb R^n$). And we need $w\cdot x_0 = d$. Once those assumptions are out of the way, we can start showing that the two descriptions give the same hyperplane (if any of these assumptions fail, then they do not describe the same hyperplane).
Choose any $\lambda_i$ for $i$ from $1$ to $n-1$, and calculate $$ w\cdot \left(x_0 + \sum_{i = 1}^{n-1}\lambda_iv_i\right) $$ and you will see that you get $d$. So any vector in $H_2$ will be in $H$.
As for inclusion the other way, take a vector $x$ such that $w\cdot x = d$. Consider the vector $x-x_0$, and write it out as a linear combination in the basis given by $w$ and the $v_i$: $$ x-x_0 = \lambda_0w + \sum_{i = 1}^{n-1}\lambda_iv_i $$ Taking the dot product with $w$ reveals that $\lambda_0 = 0$ (as $w\cdot (x-x_0) = 0$ and $w\cdot v_i = 0$ and $w\cdot w \neq 0$), so we get $$ x-x_0 = \sum_{i = 1}^{n-1}\lambda_iv_i $$ Add $x_0$ to both sides, and you're done.