I am supposed to solve a question that goes: Find the projection matrices onto the plane $x_1+x_2+3x_3+4x_4=0.$ ($x_1$,$x_2$,$x_3$ and $x_4$ are variables).
The solution goes like this: The plane is viewed as null space of matrix [1 1 3 4]. It proceeds to find the basis of the space: ${[-1\ 1\ 0\ 0],[-3 \ 0\ 1\ 0],[-4\ 0\ 0\ 1]}$.
What do they mean by saying the plane is viewed as a null space of the matrix? How did they get the basis of the space?
I have a vague idea of what null spaces and basis mean but I don't understand anything happening in the first two lines of the solution.
$\{(x_1, x_2, x_3, x_4)^T:x_1+x_2+3x_3+4x_4=0\}$ can be rewritten as
$\{(x_1, x_2, x_3, x_4)^T:\begin{bmatrix} 1 & 1 & 3 & 4\end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix}=0\}$ which is the null space of $\begin{bmatrix} 1 & 1 & 3 & 4\end{bmatrix}$.
To find its basis, from $x_1+x_2+3x_3+4x_4=0$
Let $x_2=s, x_3=u, x_4=w$, then we can write $x_1=-x_2-3x_3-4x_4=-s-3u-4w$.
That is $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix}=s\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0\end{bmatrix} +u\begin{bmatrix} -3 \\ 0 \\ 1 \\ 0\end{bmatrix} +w\begin{bmatrix} -4 \\ 0 \\ 0 \\ 1\end{bmatrix}$$