Four Dimensional intersection point How can I find the intersection point of two linearly independent 4D dimensional planes? I know that the first plane A goes through the four-dimensional points A1(1,2,3,4) A2(0,1,0,1) A3(0,1,1,0) and plane B goes through points B1(2,2,2,2) B2(4,3,2,1) B3(1,1,1,0). I know that the intersection is a point and not a line, can you explain why? I have tried considering the points as 2 matrices, but I get stuck. There is also another method to find the intersection?
 A: As noticed by SmileyCraft in the comments, the first plane is


*

*$A_1+s(A_2-A_1)+t(A_3-A_1)$
that is


*

*$x=1-s-t$

*$y=2-s-t$

*$z=3-3s-2t$

*$w=4-3s-4t$
from which we obtain


*

*$x-y=-1$

*$-6x+z+w=1$
The second plane is


*

*$B_1+s(B_2-B_1)+t(B_3-B_1)$
that is


*

*$x=2+2s-t$

*$y=2+s-t$

*$z=2-t$

*$w=2-s-2t$
from which we obtain


*

*$x-y-2z+w=-2$

*$y-3z+w=-2$
Therefore we obtain a system of four equation with four uknown.
A: Your idea of examining matrices constructed from the given points is on the right track, but you need to look at the right one. Per SmileyCraft’s comment, the first plane consists of all affine combinations $(1-\lambda-\mu)A_1+\lambda A_2+\mu A_3$ and the second $(1-\sigma-\tau)B_1+\sigma B_2+\tau B_3$. Equating these two expressions gives us a system of homogeneous linear equations in the four parameters, which we can package up into the matrix equation $$\begin{bmatrix} A_2-A_1 & A_3-A_2 & B_1-B_2 & B_1-B_3 \end{bmatrix} \begin{bmatrix}\lambda \\ \mu \\ \sigma \\ \tau\end{bmatrix} = B_1-A_1.$$ For this to have a unique solution, the $4\times4$ coefficient matrix on the left must have full rank. You can check for that by computing its determinant or reducing it to echelon form. (There are other ways, but they’re at least as much work.) If you want to find the intersection point, you can of course augment the matrix and perform Gaussian elimination in the usual way.  
Addendum: As for why the intersection of two planes can be a point, it’s basically for the same reason that there can be skew lines in $\mathbb R^3$: the additional dimensions allow more ways for things to avoid each other. To illustrate this, consider a pair of coordinate planes. Each plane is spanned by a pair of coordinate axes. With only three dimensions, there’s no way to pick two pairs of axes without there being at least one common axis in both pairs, so the intersection of two coordinate planes has to be at least that common axis—a line. With four or more dimensions, on the other hand, there are enough coordinate axes to choose two disjoint pairs of them—there’s no common axis any more. Not only that, but the only point that they have in common is the origin: All four vectors that we’ve chosen are linearly independent, so there’s no non-trivial linear combination of the first pair that can produce either of the other two vectors or any linear combination of them. To put this in terms of coordinates, coordinates of points on the $x$-$y$ plane have the form $(x,y,0,0)$ and points on the $z$-$t$ plane $(0,0,z,t)$. There’s only one point with both of these forms at once: $(0,0,0,0)$.  
Algebraically, it takes $n-2$ linear equations to specify a plane in $\mathbb R^n$. The intersection of two planes, then, is described by a system of $2(n-2)$ linear equations in $n$ unknowns. For $n=3$, this gives you a $2\times3$ coefficient matrix with nullity $\ge1$. That is, if the system has a solution at all, it has an infinite number of them and the solution space is at least a line. When $n=4$, we have a $4\times4$ coefficient matrix, which allows for a unique solution—a point. As the number of dimensions increases, the number of rows grows about twice as fast as the number of columns, so loosely speaking, it’s more and more likely that the system will be inconsistent, i.e., an arbitrary pair of planes won’t intersect at all.
