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Let $V$ be a 4-dem. vector space over $\Bbb R$, $L_1$: $x_0=0, x_2=0$ and $L_2$: $x_1=0, x_3=0$ are affine planes in $V$, $v_1 = (1, 0, 1, 1)$ and $v_2 = (1, 1, 0, 1)$ are vectors in $V$. Find an intersection of $L_1 + v_1$ and $L_2 + v_2$.

Am I right that $\forall l_1 \in L_1$ and $\forall l_2 \in L_2$ $l_1 = (0, a, 0, b), l_2 = (c, 0, d, 0)$ where $a, b, c, d$ arbitrary numbers, therefore $\forall l_{11} \in L_1 + v_1: l_{11} = (1, a, 1, b) $ and $\forall l_{21} in L_2 + v_2: l_{21} = (c, 1, d, 1)$ and then their intersection is a single point $i = (1, 1, 1, 1)$?

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If you write $l_1=(0,a,0,b)$ then $l_1+v_1=(1,a,1,b+1)$ but as $b$ can be any real number, the $+1$ doesn't really matter. You'd probably write more correctly

$$\forall l_{11}\in L_1+v_1\exists a,b\in\mathbb R:l_{11}=(1,a,1,b)$$

in order to clearly express that the $b$ here is local to this formula, and not necessarily the same $b$ you used to write down $L_1$ itself.

Anyway, your conclusion is right. One plane fixes two coordinates, the other fixes the other two coordinates, so all four coordinates are fixed as you say.

In a general $d$-dimensional space, if you intersect an $m$-dimensional affine subspace with a $n$-dimensional one, you get a $(m+n-d)$-dimensional affine space as intersection. At least if the subspaces are in general position, not parallel or some such. In this case here you have $m=n=2$ and $d=4$ so you get $2+2-4=0$ as the dimension for the intersection, a point.

You can think of the $m$ and $n$ as parameters or degrees of freedom, your $a,b,c,d$. You can think of the $d$ as a set of equations which must be satisfied by these. Four equations in four variables have a unique solution unless they happen to be linearly dependent, which they won't be for general position.