# Find the line between two planes (not using vector product).

I stumbled on to this example in my math book, this is presented before the vector product is introduced.

lets say we have an equation system like this:

$$\left( \begin{array}{3} x + 2y -3z = 3 \\ a^2x + (2a+4)y - (3a+6) = 3a +6 \\ x+ay-3z= 5-a \end{array}\right )$$

Now if you calculate the determinant you will see that the system above have a solution if:

$a \neq -1$ and $a \neq 2$

If you solve for values other than those you will see that the point $(0, -1, \frac{-5}{3})$ belongs to the three planes, independent of $a$.

So if we take the case of $a=-1$ then the planes:

$P_1 = x + 2y -3z = 3$ and $P_2 = a^2x + (2a+4)y - (3a+6) = 3a +6$ will be identical.

We are then left with the equation system like this:

$$\left( \begin{array}{3} x + 2y -3z = 3 \\ x -y-3z= 6 \end{array}\right )$$

where the line $l : (x,y,z) = (5,-1,0) + t(3,0,1)$ is the solution.

Now my question to you is how can they know that the line above is the solution?

My approach would be to find another point that satisfies the equation system and then calculate the line vector and use the $(0, -1, \frac{-5}{3})$ as the the first reference point, but how can I find the second point?

Thank you!

• Obviously you could look at the equation system $\left( \begin{array}{3} x + 2y -3z = 3 \\ x -y-3z= 6 \end{array}\right )$ and guess a solution, like $(5,-1,0)$ from there you could find a linear combination $(5,-1,0) + t(x,y,z)$ that would produce the point $(0, -1, \frac{-5}{3})$. But there should be a method that does not rely on guessing? – Lukas Arvidsson May 27 '14 at 17:09

You must just get a generalized solution for $$\begin{bmatrix} 1&2&-3&| &3 \\ 1&-1&-3&|&6 \end{bmatrix}.$$After row reduction I get $$\begin{bmatrix} 1&2&-3&| &3 \\ 0&-1&0&|&1 \end{bmatrix}.$$ So if we let $z=t$ we get $y=-1$ and $x=5+3t$, and therefore we have the general solution $(x,y,z)=(5+3t,-1,t)=(5,-1,0)+t(3,0,1)$.