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

  1. Nov 1, 2008 #1
    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 [tex]a=\frac{9}{16}t, b=-\frac{1}{16}t, c= \frac{5}{16}t, d=t[/tex] "

    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)

    Thanks in advance!
     
  2. jcsd
  3. Nov 3, 2008 #2

    HallsofIvy

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    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
    [tex]\left[\begin{array}{ccccc}0 & 2 & -1 & 1 & -1 \\ 2 & 3 & 1 & 1 & 0 \\ 3 & -1 & 1 & 1 & 0\end{array}\right][/tex]
     
    Last edited: Nov 3, 2008
  4. Nov 3, 2008 #3
    Am I correct in thinking the Gaussian elimination would be the best choice, or is there a more efficient way I'm missing?
     
  5. Nov 3, 2008 #4
    I recommend you instead use the point-normal form instead.

    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
     
  6. Nov 4, 2008 #5
    Awesome, Maze! But where does the 1/16 come from then?
     
  7. Nov 4, 2008 #6

    HallsofIvy

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    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.
     
  8. Nov 4, 2008 #7
    Yes! Doing the cross product I obtained (9,1,-5), lacking the denominator I need.
     
  9. Nov 4, 2008 #8
    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.
     
  10. Nov 4, 2008 #9
    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...
     
  11. Nov 4, 2008 #10
    Maybe it will be easier to help if you can post your work so far.
     
  12. Nov 4, 2008 #11
    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
     
  13. Nov 4, 2008 #12
    Recheck the determinants,
    [tex]det(\begin{matrix}1 & 0 \\ -4 & 1\end{matrix}) = 1*1-(-4*0) = 1[/tex]
     
  14. Nov 4, 2008 #13
    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.
     
  15. Nov 4, 2008 #14
    Yeah ok with p1 = (1,2,-1) that looks good. So whats the problem then?
     
  16. Nov 4, 2008 #15
    Is there a problem with doing it my way, as I do not get the 1/16th still... or am I forgetting something?
     
  17. Nov 4, 2008 #16
    Oh, i see. Just recall

    ax + by + c = 0

    is the same as

    ax/16 + bx/16 + c/16 = 0
     
  18. Nov 4, 2008 #17
    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.
     
  19. Nov 4, 2008 #18
    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.
     
  20. Nov 10, 2008 #19
    Thanks again, maze.

    If all 3 points are on the plane, what makes you choose (x-p1) ?
     
  21. Nov 10, 2008 #20
    You could choose any point on there. Give it a try for p3 and see what happens!
     
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