[Math] Finding the line of intersection of 2 planes by row reducing the linear system matrix

linear algebramatrices

Problem Statement: Find the plane that passes through the point (-1, 2, 1) and contains the line of intersection of the planes:

$$x+y-z = 2$$

$$2x – y + 3z = 1$$

I understand there is a means of solving this with the cross product – but I am interested in whether or not I can solve this by using a matrix to represent the linear system.

$$ A =
\left[\begin{array}{rrr|r}
1 & 1 & -1 & 2 \\
2 & -1 & 3 & 1
\end{array}\right]
$$

By row reducing the matrix we find:

$$ RREF(A) =
\left[\begin{array}{rrr|r}
1 & 0 & -2/3 & 1 \\
0 & 1 & -5/3 & 1
\end{array}\right]
$$

Therefore, the system becomes:

$$x – 2/3z = 1 $$
$$y – 5/3z = 1$$

And by parametrizing $z = t$, and solving the system for $x$ and $y$:

$$x = 2/3t + 1$$
$$y = 5/3t + 1$$
$$z = t$$

Problem is, this resulting system does not actually match the line of intersection. Is there something I'm neglecting or doing wrong?

Show[
 ContourPlot3D[
  {x + y - z == 2, 2 x - y + 3 z == 1}, {x, -10, 10}, {y, -10, 
   10}, {z, -10, 10},
  Boxed -> False, Axes -> True, AxesOrigin -> {0, 0, 0}
  ],
 ParametricPlot3D[
  {(2/3) t + 1, (5/3) t + 1, t},
  {t, -10, 10}
  ]]

pic

Best Answer

You say

$RREF(A) = \left[\begin{array}{rrr|r} 1 & 0 & -2/3 & 1 \\ 0 & 1 & -5/3 & 1 \end{array}\right]$

However,

$RREF(A) = \left[\begin{array}{rrr|r} 1 & 0 & 2/3 & 1 \\ 0 & 1 & -5/3 & 1 \end{array}\right]$

(I just switched the -2/3 to 2/3.) The first row of RREF(A) is just the first row of A minus the second row of RREF(A). I am not sure if this solves it completely, but that could be it.

Related Question