# Line of Intersection of 2 Planes

I'm trying to understand how to find line of intersection between 2 planes. In almost every explanation I've seen, it involves removing one of the unknowns by substituting it with 0.

For example in this video https://www.youtube.com/watch?v=q2FO675dmyM

For the planes:
x + 2y - 4z = 16
2x - y + 3z = 6

Substituting x = 0,
We get the equations
2y - 4z = 16
2y + 6z = 12


And eventually we can solve the equation.

But what if we cannot substitute x = 0?

Consider the planes:
x + y = 8
x = 5


In this case, the line of intersection will never touch x = 0 or y = 0. I know that with this equation, we can already easily solve this question as there is only 2 unknowns (technically 1 unknown since x = 5).

But I'm just curious if it is possible for such a case to occur with 3 unknowns. Where we have 3 unknowns but the line of intersection will never pass through one of the axis.

Rather than assign the value $$0$$ to $$x$$, for the reasons you mention, (or $$y$$ or $$z$$, for that matter), it is better to eliminate one of the letters from the two equations first.

So, for example, taking the equations you mention:$$x+2y-4z+16$$ and $$2x-y+3z=6$$

If we decide to eliminate $$y$$ for example we get $$5x+2z=28$$ Whereupon, $$5x=28-2z=\lambda$$

We can then get each of $$x$$ and $$y$$ and $$z$$ in terms of the parameter $$\lambda$$, which gives us an equation of the line.

We are trying to find nonzero $$x_1, x_2, y_1, y_2, z_1, z_2$$ and real $$c_1, c_2$$ such that for the system of equations

$$$$\begin{cases} x_1x+y_1y+z_1z=c_1 \\ x_2x+y_2y+z_2z=c_2 \end{cases}$$$$

such that this is equivalent to $$x=a$$ for some nonzero $$a$$.

This notation is rather clunky, so we can use matrices to represent this:

$$\left(\begin{array}{rrr|r} x_1 & y_1 & z_1 & c_1 \\ x_2 & y_2 & z_2 & c_2 \end{array}\right)$$

Now we can begin using Gaussian Elimination to solve this system of equations.

We first divide the first row by $$x_1$$:

$$\left(\begin{array}{rrr|r} 1 & \frac{y_1}{x_1} & \frac{z_1}{x_1} & \frac{c_1}{x_1} \\ x_2 & y_2 & z_2 & c_2 \end{array}\right)$$

Now we get rid of the $$x_2$$ term by subtracting $$x_2$$ copies of row 1 from row 2:

$$\left(\begin{array}{rrr|r} 1 & \frac{y_1}{x_1} & \frac{z_1}{x_1} & \frac{c_1}{x_1} \\ 0 & y_2-\frac{y_1x_2}{x_1} & z_2-\frac{z_1x_2}{x_1} & c_2-\frac{c_1x_2}{x_1} \end{array}\right)$$

Now divide row 2 by $$y_2-\frac{y_1x_2}{x_1}$$:

$$\left(\begin{array}{rrr|r} 1 & \frac{y_1}{x_1} & \frac{z_1}{x_1} & \frac{c_1}{x_1} \\ 0 & 1 & \frac{z_2-\frac{z_1x_2}{x_1}}{y_2-\frac{y_1x_2}{x_1}} & \frac{c_2-\frac{c_1x_2}{x_1}}{y_2-\frac{y_1x_2}{x_1}} \end{array}\right)$$

Finally subtract row 1 by $$\frac{y_1}{x_1}$$ copies of row $$2$$:

$$\left(\begin{array}{rrr|r} 1 & 0 & \frac{z_1}{x_1}-\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)} & \frac{c_1}{x_1}-\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)} \\ 0 & 1 & \frac{z_2-\frac{z_1x_2}{x_1}}{y_2-\frac{y_1x_2}{x_1}} & \frac{c_2-\frac{c_1x_2}{x_1}}{y_2-\frac{y_1x_2}{x_1}} \end{array}\right)$$

Let's look at the first row, which equivalently is:

$$x+\left(\frac{z_1}{x_1}-\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}\right)z=\frac{c_1}{x_1}-\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}$$

Recall that we want $$x=a$$ for nonzero $$a$$, so we need:

$$\frac{z_1}{x_1}-\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}=0$$ $$\frac{c_1}{x_1}-\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}\neq0$$

Both look awfully complicated so we can clean up both equations.

Working on the first equation:

$$\frac{z_1}{x_1}=\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}$$ $$z_1=\frac{y_1\left(z_2-\frac{z_1x_2}{x_1}\right)}{y_2-\frac{y_1x_2}{x_1}}$$ $$z_1\left(y_2-\frac{y_1x_2}{x_1}\right)=y_1\left(z_2-\frac{z_1x_2}{x_1}\right)$$ $$z_1(y_2x_1-y_1x_2)=y_1(z_2x_1-z_1x_2)$$

As for the second equation:

$$\frac{c_1}{x_1}\neq\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{x_1\left(y_2-\frac{y_1x_2}{x_1}\right)}$$ $$c_1\neq\frac{y_1\left(c_2-\frac{c_1x_2}{x_1}\right)}{y_2-\frac{y_1x_2}{x_1}}$$ $$c_1\left(y_2-\frac{y_1x_2}{x_1}\right)\neq y_1\left(c_2-\frac{c_1x_2}{x_1}\right)$$ $$c_1(y_2x_1-y_1x_2)\neq y_1(c_2x_1-c_1x_2)$$

Combining,

$$$$\begin{cases} z_1(y_2x_1-y_1x_2)=y_1(z_2x_1-z_1x_2) \\ c_1(y_2x_1-y_1x_2)\neq y_1(c_2x_1-c_1x_2) \end{cases}$$$$

Considering there are $$8$$ different constants and $$2$$ equations, there should be infinitely many $$x_1,x_2,y_1,y_2,z_1,z_2,c_1,c_2$$ satisfying both equations.

An example of such a system of equations is:

$$$$\begin{cases} x+y+z=2 \\ x+2y+2z=3 \end{cases}$$$$

Which is equivalent to

$$$$\begin{cases} x=1 \\ y+z=1 \end{cases}$$$$