# Equation of a plane (General and Point-Normal Form)

1. Nov 1, 2008

Hey all,
I'm trying to figure out an example from my Linear Algebra Book:

The question is as follows:
Find the equation of the plane passing through the points P1(1,2,-1), P2(2,3,1) and P3(3,-1,2)

It goes on to say it must satisfy this system, which is fine:

1 + 2b - c + d = 0
2a + 3b + c + d = 0
3a - b + 2c + d = 0

Then it just says, "solving for this system gives $$a=\frac{9}{16}t, b=-\frac{1}{16}t, c= \frac{5}{16}t, d=t$$ "

To solve this system, would one include the d's in the matrix or leave them out?
What method would you recommend for solving this? (and then I can give it a shot)

2. Nov 3, 2008

### HallsofIvy

You forgot to mention that you are assuming the equation of the plane (or any plane) can be written ax+ by+ cz+ d= 0! I wondered where a, b, c, and d came from!

Yes, of course you have to include d! You have to include all the variables. The equations you give are equivalent to
2b- c+ d= -1
2a+ 3b+ c+ d= 0
3a- b+ 2c+ d= 0
The augmented matrix for that system of equations is
$$\left[\begin{array}{ccccc}0 & 2 & -1 & 1 & -1 \\ 2 & 3 & 1 & 1 & 0 \\ 3 & -1 & 1 & 1 & 0\end{array}\right]$$

Last edited by a moderator: Nov 3, 2008
3. Nov 3, 2008

Am I correct in thinking the Gaussian elimination would be the best choice, or is there a more efficient way I'm missing?

4. Nov 3, 2008

### maze

To find the normal, you can take the cross product n = (p2-p1)X(p3-p2). Then the condition that the normal is perpendicular to vectors on the plane means that (x-c)*n = 0 for points x and c on the plane.

Here is a diagram:
http://img259.imageshack.us/img259/7929/pointnormalformpy3.png [Broken]

Last edited by a moderator: May 3, 2017
5. Nov 4, 2008

Awesome, Maze! But where does the 1/16 come from then?

6. Nov 4, 2008

### HallsofIvy

Have you tried
1) Solving the system of equations?
2) Reducing the matrix I gave you?
3) Taking the cross product as Maze suggested?

Doing any of those things will show you where the "1/16" came from. The point is that you have three equations in 4 unknown coefficients. You can solve for any three in terms of the other one. Here they chose to solve for a, b, c in terms of d and use d as the parametr.

Actually, it doesn't matter. If you define s to be "t/16", then, a= 9s, b= -s, c= 5s and d= 16s.

7. Nov 4, 2008

Yes! Doing the cross product I obtained (9,1,-5), lacking the denominator I need.

8. Nov 4, 2008

### maze

Hmm. heres what i get
p1=(1,2,1), p2=(2,3,1), p3 = (3,-1,2)

p2-p1=(1,1,0), p3-p2=(1,-4,1)

n = (p2-p1)X(p3-p2) = i(1-0) + j(0-1) + k(-4-1) = (1,-1,-5)

0 = n*(x-p1) = (1,-1,-5)*(x1-1, x2-2, x3-1) = x1-1-x2+2-5*x3+5
0 = x1 - x2 - 5*x3 + 6

This checks out with the 3 points, so it should be good.

9. Nov 4, 2008

Hey Maze,
sorry, I'm not following your last two steps.

What I did is set it up as a 2x3 matrix and solved for the three components.
I used the easy method of "crossing" out a column and then finding the three determinants.

This gives me the three numbers from before, but not the 1/16th...

10. Nov 4, 2008

### maze

Maybe it will be easier to help if you can post your work so far.

11. Nov 4, 2008

Sure,

I'm just using a trick from my linear algebra book:

taking the cross product of a 2x3

1 1 0
1 -4 1

The trick is, to get the x portion, cross out the first column and take the determinant. For the y, cross out the second column, for the z, cross out the third.

The three determinants I get are 9, 1, -5

12. Nov 4, 2008

### maze

Recheck the determinants,
$$det(\begin{matrix}1 & 0 \\ -4 & 1\end{matrix}) = 1*1-(-4*0) = 1$$

13. Nov 4, 2008

Wait, sorry - I accidentally used your value. The first row should be 1 1 2.

(P1 is 1,2, -1)

Using that, I get the values I showed.

14. Nov 4, 2008

### maze

Yeah ok with p1 = (1,2,-1) that looks good. So whats the problem then?

15. Nov 4, 2008

Is there a problem with doing it my way, as I do not get the 1/16th still... or am I forgetting something?

16. Nov 4, 2008

### maze

Oh, i see. Just recall

ax + by + c = 0

is the same as

ax/16 + bx/16 + c/16 = 0

17. Nov 4, 2008

I'm confused...

when I wrote the solutions with 1/16th in my first post - that came from an unsolved answer in the book. I do not know how its obtained solving for my determinants,etc.

18. Nov 4, 2008

### maze

You got the normal by cross product: n = (9,1,-5)

Now since the normal is perpendicular to any vector parallel to the plane, and since (x-p1) is parallel to the plane if x is on the plane, we have

0 = n*(x-p1)

0 = (9,1,-5)*((x1,x2,x3)-(1,2,-1)) = (9,1,-5)*(x1-1,x2-2,x3+1)
0 = 9*x1 - 9 + x2 - 2 - 5*x3 - 5
0 = 9*x1 + x2 - 5*x3 - 16

So there you go, thats the equation for the plane. You could divide the whole thing by 16 to get the results the book talks about.

19. Nov 10, 2008

Thanks again, maze.

If all 3 points are on the plane, what makes you choose (x-p1) ?

20. Nov 10, 2008

### maze

You could choose any point on there. Give it a try for p3 and see what happens!