Angle of two planes in $\mathbb E_4$

Question

I have two planes (given in parametric form) in $\mathbb E_4$: $\alpha$: (7,3,5,1) + t(0,0,1,0) + s(3,3,0,1) and $\beta$: (1,5,4,1) + r(0,0,0,-1) + p(2,0,0,1), and I have to find angle between them. I start with search for crossing of their directions by solving matrix of their direction vectors: t(0,0,1,0) + s(3,3,0,1) = r(0,0,0,-1) + p(2,0,0,1): \begin{matrix} 0 & 3 & 0 & -2\\ 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & -1 \end{matrix} Then I should find the crossing between direction vectors of $\alpha$ and $\beta$ and result of that. However, the matrix can be changed into standart base of $\mathbb E_4$, and I am not sure what that means. Can somebody explain to me, how should I proceed (in this paradigm, please)?

Answer 1

I didn't check the numbers in details, just shortly: If the directional vectors of the two 2-planes (i.e. $(0,0,1,0), \ldots, (2,0,0,1)$) span a 3-dimensional vector subspace $A$ of $\mathbb{E}_4$, then the angle between them is defined to be the angle of the normal vectors to those planes (normal in $A$). If those 4 vectors are independent, I don't see a reasonable definition of what the angle should be

(You could find a minimal angle between a vector in $\alpha$ and in $\beta$, but I don't think it is something standard).