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}
]]
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.