Parametric form of a plane

Can you please explain to me how to get from a nonparametric equation of a plane like this:

$$x_1−2x_2+3x_3=6$$

to a parametric one. In this case the result is supposed to be

$$x_1 = 6-6t-6s$$ $$x_2 = -3t$$ $$x_3 = 2s$$

Many thanks.

Welcome to math.stackexchange!

A plane can be defined by three things: a point, and two non-colinear vectors in the plane (think of them as giving the plane a grid or coordinate system, so you can move from your first point to any other using them).

So first, we need an initial point: since there are many points in the plane, we can pick randomly. I'll just take $x_1=6,x_2=0$ so that $x_3=0$ and we see that the point $(6,0,0)$ solves the equation.

Now I need two vectors in the plane. I can do this by finding two other points in the plane, and subtracting them from this one (the difference of two vectors points from one to the other, so if both points are in the plane their difference will point along it). I'll take the points $(0,-3,0)$, and $(0,0,2)$. Notice the simple construction of all my points: set two variables to zero and find out what the third one should be. You can almost always do this, and it's probably the easiest way to go.

So my vectors are going to be these two points minus the original one I found. $$(0,-3,0)-(6,0,0)=(-6,-3,0)$$ $$(0,0,2)-(6,0,0)=(-6,0,2)$$ Now any vector in the plane, when scaled, is still in the plane. So I can define my plane like this: $$(6,0,0)+(-6,-3,0)t+(-6,0,2)s$$

I.e. start at the first point, and move $t$ amount in one direction and $s$ amount in another, where $t$ and $s$ range over the real numbers, so they cover the whole plane. Note that each of the scaled vectors, when plugged into the equation, give $0$. So for any point here, we're doing $6+0+0=6$, which solves the original equation. Splitting this up in terms of components $(x_1,x_2,x_3)$ instead of points, we get $$x_1=6-6t-6s$$ $$x_2=-3t$$ $$x_3=2s$$

There are infinitely many other parameterizations that could have worked, so your answer could look completely different while still being completely correct. But this is probably the logic they used, in case you were wondering.

• Thank you very much indeed for your great answer and for welcoming me here! I hope I'll be able to learn how to write all the tags properly so you won't need to correct my questions again. Thanks. Jun 1 '12 at 14:29
• Latex is quite easy to pick up. We're fortunate enough to have the folks over at tex.stackexchange.com helping us out if we need it. For this question there were only 3 things I did: surrounding things with single dollar signs makes them look nice inline. Using double dollar signs gives things their own line. And using an underscore within those dollar signs (called 'math mode') makes subscripts. If you want to see how any latex was made, just right click on it and go to Show math as -> tex commands. Jun 1 '12 at 14:33
• @RobertMastragostino Is there a mistake here? Shouldn't the third point be (0, 0, 2) instead of (0, 2, 0)? Nov 15 '13 at 18:55
• @AndyG Yes, it seems it should be. I'll update, thanks Nov 18 '13 at 4:17
• What a beautiful answer. Masterful exposition! Sep 1 '17 at 16:07

One way to do it is to let $x_1 = t$ and $x_2=s$ and then solve for $x_3$.

• but how can i get to that result i wrote? Jun 1 '12 at 14:18
• Look at $x_2$ and $x_3$ in your case, can you then somehow get $x_1$?
– Sean
Jun 1 '12 at 14:20
• Many thanks for writing me back. But if I just follow what you wrote I get x3 = 6/3 +2/3 s- t/3 Jun 1 '12 at 14:25
• Yip that's correct. This is equivalent to your solution in your question.
– Sean
Jun 1 '12 at 14:31
• All right, thanks anyway! Jun 1 '12 at 14:32

There is more than one way to write any plane is a parametric way. To write a plane in this way, pick any three points $A$, $B$, $C$ on that plane, not all in one line. Then $$f(s, t) = A + (B-A)s + (C-A)t$$