Intersection of two planes in $\mathbb{R}^4$

Question

I have two planes which i have to intercept, but my answer isn't correct (i think)

Plane I $= (-14, 8, 3, 3) + r(3,3−,3,0) + s(1, −1, −3, −1)$

Plane II $= (-7, 1, 3, 6) + u(-2, 0, -2, -2) + v(-3, 1, -2, 3)$

then i mount this matrix and use Gauss-Jordan method \begin{equation} \begin{bmatrix} -3 & 1 & 2 & 3 & |&7 \\ 3 & -1 & 0 & -1 & |&-7 \\ -3 & -3 & 2 & 2 & |&0 \\ 0 & -1 & 2 & -3 &|& 3 \end{bmatrix} \end{equation}

It returned $x_1 =2$, but haven't this option to chose. Im wrong or this question no make sense? im so confused lol.

Answer 1

I think the matrix is incorrect, please refer below

$$-14 + 3r + s = -7 -2u -3v \implies 3r + s + 2u +3v = 7$$

$$8 -3r -s = 1 +0u +v \implies -3r -s +0u -v = -7$$

$$3 + 3r -3s = 3 - 2u-2v \implies 3r -3s + 2u + 2v = 0$$

$$3 + 0r - s = 6 -2u +3v \implies 0r-s+2u-3v = -3$$

Hence the matrix looks like

$$\begin{bmatrix}3 & 1 & 2 & 3 \\ -3 & -1 & 0 &-1 \\ 3 & -3 & 2 & 2 \\ 0 & -1 &2 &-3\end{bmatrix} \begin{bmatrix}r \\ s \\ u \\ v\end{bmatrix} = \begin{bmatrix}7 \\ -7 \\ 0 \\ 3\end{bmatrix}$$

Now when you solve this matrix, you will get values of $r,s,u,v$ for which the planes intersect. The next step would be to input either in one plane or the other the corresponding variable values to get the line of intersection

Answer 2

Let's talk about how the method works. You assume there exists a point $P$, which lie both in plane I and plane II: $$ \vec P = \vec a_1+\vec b_1 r+\vec c_1 s=\vec a_2+\vec b_2 u+\vec c_2 v, $$

thus you make it an equation: $$ \vec b_1 r+\vec c_1 s-\vec b_2 u-\vec c_2 v = \vec a_2-\vec a_1 $$

which can be written is matrix form: $$ \begin{pmatrix} b_{11}&c_{11}&b_{21}&c_{21} \\ b_{12}&c_{12}&b_{22}&c_{22} \\ b_{13}&c_{13}&b_{23}&c_{23} \\ b_{14}&c_{14}&b_{24}&c_{24} \\ \end{pmatrix}\begin{pmatrix} r\\s\\-u\\-v \end{pmatrix} = \begin{pmatrix} a_{21}-a_{11}\\a_{22}-a_{12}\\a_{23}-a_{13}\\a_{24}-a_{14} \end{pmatrix} $$ To solve this, you use the Gauss method or any other.

But note that, you are solving for $r,s,u,v$, not the coordinate of the point $P$.