# Visualizing a plane

#### DWill

I'm having trouble visualizing what the graph of the plane x + y + z = 2 looks like. It is easier for me to graph equations of 2 variables. Also, I thought equations of 2 variables are "planes", while equations of 3 variables are of a solid? Is there any steps to follow or something to graph these equations of 3 variables?

#### Defennder

Homework Helper
I'm having trouble visualizing what the graph of the plane x + y + z = 2 looks like. It is easier for me to graph equations of 2 variables.
Trying graphing for selected values of z. Let z=0, z=-1, z=1 for example. Then visualise a plane connecting those lines for those respective z values in 3D.

Also, I thought equations of 2 variables are "planes", while equations of 3 variables are of a solid? Is there any steps to follow or something to graph these equations of 3 variables?
There are lines and planes in 3D as well which may require 3 variables to represent. You can't, for example, represent the equation of a line in 3D by a single equation. The steps you should follow in visualising are about graphing what the "level sets" or "level curves" would look like for a specific value of z,x or why.

#### HallsofIvy

Homework Helper
It requires 3 numbers to designate point in 3 dimensions (that's why it is 3 dimensions!) If you write a single equation connecting those three numbers, then, given any two you could solve for the third. That's why a single equation, in 3 dimensions, represents a two dimensional object such as a plane.

One way you can "visualize" a plane is to use the "intercepts" like in 2 dimensional graphing of a line. Any point on the z-axis, for example, has both x and y equal to 0. In your example, x+ y+ z= 2, if x and y are both equal to 0, z= 2. The plane must contain the point (0, 0, 2). In fact, for that equation it is easy to see that the other two intercepts are (2, 0, 0) and (0, 2, 0). Those three points form a triangle and the portiona of the plane in the first "octant" (where the 3 variables are all positive) is the interior of that triangle. The plane itself "lies on" that triangle.

Defennnder's method, of looking at different of z, also works very nicely for non-linear functions in which the surfaces are not planes. It gives the "countour maps" just like geodesic maps of mountains that show altitude.

#### maze

I recommend you calculate the plane normal vector for easier visualization.

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