Suppose $S$ is an affine subspace of $\mathbb{R}^n$ of dimension $n-1$ which contains the point $P=(p_1,\ldots,p_{n-1},p_n)$, if $X=(x_1,\ldots,x_{n-1},x_n)$ is a point of $\mathbb{R}^n$ is it "right" to say: $$X \in S \iff x_n=p_n+ \langle (tan \ \theta_1,\ldots,tan \ \theta_{n-1}),(x_1 - p_1,\ldots ,x_{n-1}-p_{n-1})\rangle$$ where $\theta_i$ denotes the angle between $S$ and the $x_i$ axis?
I'm asking if it's "right" because I'm not really sure if it makes sense to talk about the angle between $S$ and the $x_i$ axis when we're dealing with $n>3$ since I'm having trouble trying to visualize for example a 3-dimentional subspace of $\mathbb{R}^4$ cutting at some point one of the axis. I'm just trying to generalize the equation of a line in $\mathbb{R}^2$ or a plane in $\mathbb{R}^3$ but I have to admit I don't know much of affine spaces, I took a two-semesters course on linear algebra but it didn't cover that topic.
Thanks
Informally an affine subspace is a linear subspace (which necessarily contains the origin) possibly shifted by some fixed vector.
So we can indeed split the the point $x$ into a fixed position vector $p$ and a vector from a linear subspace $v$: $$ x = p + v $$ For a slope $m$ of a function of one variable we have $$ \tan \theta = \frac{m}{1} = \frac{\Delta y}{\Delta x} \to y'(x) $$ Your vector $t = (\tan \theta_1, \dotsc, \tan \theta_n)$ seems to be a gradient $\DeclareMathOperator{grad}{grad}\grad f = (\partial f / \partial x_1, \dotsc, \partial f / \partial x_n)$. For a plane it would be a normal vector of that plane.
Your equation, which I interpret as $$ x_n = p_n + \alpha t + \beta (x-p) $$ for some scalar values $\alpha$, $\beta$ does not make sense to me. It ought to be $$ x_n = p_n + \alpha v_1 + \beta (x-p) $$ with $v_1$ being another vector within the plane, next to $x - p$, but linear independent from it.
You can make use of a normal vector like this:
The affine subspace $S$ is an affine hyperplane (subspace of dimension $n-1$), thus its points $x \in S$ fulfill an equation $$ a \cdot x = b $$ for some non zero vector $a$ and some scalar $b$ which can be divided by $\lVert a \rVert$ on both sides to give the equation $$ n \cdot x = d $$ for a unit normal vector $n$ of $S$ and distance $\lvert d\rvert$ of $S$ to the origin.
The angle between $n$ and $e_i$ by definition of the scalar product is $$ n_i = n \cdot e_i = \lVert n \rVert \lVert e_i \rVert \cos \angle(n, e_i) = \cos \angle(n, e_i) $$ which is the $i$-th component of $n$.
The advantage seems to me that there are only two unit normal vectors $\pm n$ to a hyperplane (two sides), while there are infinite many vectors within (parallel to) the plane and therefore only one or two angles $\pm n_i$ exist to choose from.