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

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

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:


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:

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


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

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:

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

Which is equivalent to

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


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