# Finding basis from intersection of two planes

I want to know the general procedure of finding basis of the intersection of two planes.

Let \begin{align} S = \{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 | a_1 x_1 + a_2 x_2 + a_3 x_3 + a_4 x_4 =0 \} \cap \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 | b_1 x_1 + b_2 x_2 + b_3 x_3 + b_4 x_4 =0\} \end{align} Then how one can find a basis for $$S$$?

It seems for me it is nothing but finding null space of \begin{align} A = \begin{pmatrix} a_1 & a_2 & a_3 & a_4 \\ b_1 & b_2 & b_3 & b_4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{align} Am I right? How one can prove this?

Can we generalized to this with many planes? [I mean for example introduced $$c_1x_1 + c_2 x_3 + c_3 x_3 + c_4 x_4 =0$$ and do the similar process.]

If $$x\in \mathbb{R}^4$$ must lie in $$S$$, then it must satisfy $$\sum_i a_i x_i = 0$$ and $$\sum_i b_i x_i = 0$$ (lying on both planes). Thus, it must be in the null space of '$$A$$' you mentioned. The question reduces to finding the null space.
Assuming the rank of $$A$$ is two (in general, this is the case), its null space has a dimension of 2. Hence, any vector belonging to the null space can be completely defined by two variables. Let us choose $$x_3$$ and $$x_4$$ as these 'free' variables. Since $$Ax=0$$, we have $$a_1 x_1 + a_2 x_2 = -x_3(a_3) - x_4(a_4)\\ b_1 x_1 + b_2 x_2 = -x_3(b_3) - x_4(b_4)$$ Given values of $$x_3$$ and $$x_4$$, one can obtain $$x_1$$ and $$x_2$$ from the above equations. To get a basis, one can arbitrarily chose a set of $$(x_3,x_4)$$ pair. A simple choice is the set $$\{(1,0),(0,1)\}$$. The $$x_1$$ and $$x_2$$ values corresponding to these will fully define the basis vectors.