Finding line of intersection between two planes by solving a system of equations

Question

Ok, so the question is to find the line of intersection between two planes, given their equations.

$x+3y+2z=4$

$x-y-z=4$

I know there's the way of using the vector perpendicular to both normals of the planes as the direction vector of the line of intersection and then finding a specific point on the line.

But I wanted to find the line of intersection by solving the system- so taking one variable as a parameter and putting the two other variables in terms of that variable. So I just randomly chose x to act as the parameter:

$3y+2z=4-x$

$-y-z=4-x$

$y=x+4-z$

and then I substituted this y to $3y+2z=4-x$

$3x+12-3z+2z=4-x$

$z=8+4x$

then

$y=x+4-8-4x$

$y=-4-3x$

and I put x=t and so

$x=t, y=-4-3t, z=8+4t$

But this doesn't correspond with the answer which is $x=4+t, y=-3t, z=4t$. And I think it's because I took the wrong variable as the parameter..? Is there like a set rule for choosing which variable to be the parameter?

Answer 1

Multiplying your second equation by $4$ and adding to the first we get $$5x=20+y$$ so $$x=4+\frac{y}{5}$$ we $y=5t$ we get $$x=4+t$$ and $t$ is a real number.

Answer 2

Having

$$ a_1x+b_1y+c_1z= d_1\\ a_2x+b_2y+c_2z=d_2 $$

or

$$ \left[ \begin{array}{ccc} a_1 & b_1 & c_1\\ a_2&b_2 & c_2 \end{array} \right] \left[ \begin{array}{c} x\\ y\\ z \end{array} \right] = \left[ \begin{array}{c} d_1\\ d_2 \end{array} \right] $$

Now choosing an invertible $2\times 2$ submatrix in $\left[ \begin{array}{ccc} a_1 & b_1 & c_1\\ a_2&b_2 & c_2 \end{array} \right]$

for instance $\left[ \begin{array}{cc} a_1 & c_1\\ a_2 & c_2 \end{array} \right]$

we rearrange the system as

$$ \left[ \begin{array}{cc} a_1 & c_1\\ a_2 & c_2 \end{array} \right] \left[ \begin{array}{c} x\\ z \end{array} \right] = \left[ \begin{array}{c} d_1\\ d_2 \end{array} \right] -y \left[ \begin{array}{c} b_1\\ b_2 \end{array} \right] $$

and then

$$ \left[ \begin{array}{c} x\\ z \end{array} \right] = \left[ \begin{array}{cc} a_1 & c_1\\ a_2 & c_2 \end{array} \right]^{-1} \left( \left[ \begin{array}{c} d_1\\ d_2 \end{array} \right] -t \left[ \begin{array}{c} b_1\\ b_2 \end{array} \right] \right)\\ y = t $$