How to find the line of interesection of 2 planes (WITHOUT cross products)

Question

I'm looking to find the line of intersection between x + y - z = 4 and -x - y - z = 1.

I've tried the methods suggested in other stackexchange posts, as well as following online tutorials, but they dont seem to work based on the setup of the equations. What happens is everything cancels out leaving me with z = -2.5 or something, and I'm just left confused. Any help to find this line in vector form?

Answer 1

We have a system of two equations in 3 variables, in matrix form

$$\begin{bmatrix} 1 & 1 & -1 & | & 4\\ -1 & -1 & -1 & | & 1\end{bmatrix}$$

which after adding the equations becomes

$$\begin{bmatrix} 1 & 1 & -1 & | & 4\\ 0 & 0 & -2 & | & 5\end{bmatrix}$$

It shows that $y$ is the "free variable" and we can solve the rest in terms of it: set $y=t \in \Bbb R$, then the last equation is simple and just gives $z=-\frac52$, backsubstituting $y=t$ and $z = -\frac{5}{2}$ in the first equation gives $$x+t + \frac{5}{2}=4 \iff x = \frac{3}{2}- t$$

which gives the line of intersection as

$$\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}\frac{3}{2}\\0\\-\frac{5}{2}\end{pmatrix} + t \begin{pmatrix}-1\\1\\0\end{pmatrix}$$

Answer 2

You're off to a good start!

Now, put it back into one of the equations you had, and you'll find $x+y=1.5$ and $z=-2.5$ give the line you need.

Answer 3

You get $z=-2.5$ and then plugging this back in $x+y=1.5$, which is just another way of intersecting two planes.

However take a point on this line, say with $x=t, y=1.5-t, z=-2.5$. Can you see how to put that into vector form?

Answer 4

All you need is two points on the intersection of your planes.

Therefor let $ x=0$ in your equations and solve for $y$ and $z$ to get the first point.

Then let $y=0$ and find $x$ and $z$ for the second point.

Now that you have two points write the equation of the line passing through these points.

I found two points $(0,3/2,-5/2)$ and $(3/2,0,-5/2)$

with the equation of the line being $$x=\frac {3}{2} t, y=\frac {3}{2} (1-t), z=-5/2$$

Of course you can assign other values to $x$ or $y$ to find your points but usually $0$ is a good choice.