# Finding the line of intersection of 2 planes by row reducing the linear system matrix

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}
]]


-
Hint: whenever you find a "solution" that does not satisfy the original equations (and you did not perform any steps that possibly introduce spurious solutions), work back through your steps to find the first point where it does satisfy your intermediate equations; your error must have been just before. – Marc van Leeuwen Sep 12 '12 at 5:16

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.

-
Haha, turns out I can't read my own writing. Thanks for the help! EDIT: Can't select the answer for a few minutes. – JT Cho Sep 12 '12 at 4:59
Yea I figured it would be something like that =P – mathguy Sep 12 '12 at 4:59