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Question about determinate form of plane

  1. Jun 4, 2013 #1
    1. The problem statement, all variables and given/known data

    I was just curious about determinate form of a plane if anyone has any info.
    If you check out number 18 on wolfram I was pondering about this form. http://mathworld.wolfram.com/Plane.html
    2. Relevant equations

    3. The attempt at a solution

    I would like to know why if you had a vector a = <x,y,z> When in the form in number 18 you have a 1? Also I read somewhere else that if you were to put any of the points say <x1,y1,z1> In for the ,x,y,z the determinate has two rows that are the same so it evaluates to 0? If you have a good knowledge of this form then spill your beans about this I would appreciate it.Thanks
  2. jcsd
  3. Jun 4, 2013 #2


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    You can substitute any nonzero number for 1 in the 4x4 matrix form and it will still work since the final value is zero. The 3x3 form of the determinant is more geometrically informative. It just says that the difference between <x,y,z> and one of the points in your plane is a linear combination of the vector differences between the other two vectors in formed by difference of points in your plane. That's closer to the geometric feel of what a plane is. Expressing it as a 4x4 determinant is more of a trick than anything else.
  4. Jun 4, 2013 #3
    So if you had to express a general point say x,y,z and you had three other vectors you would have to do that with a 4x4 matrix correct? Thanks for the reply.

    I guess I still don't understand why you need a 1 or any other number as long as the column is the same? (maybe the last column could be all 6's or w.e) Why could you not have a 4x3 or something?
    Last edited: Jun 4, 2013
  5. Jun 4, 2013 #4


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    No, you don't HAVE to. That was my point. There's a 3x3 matrix following that's exactly equal to the 4x4. But that just expresses that <x,y,z>-<x1,y1,z1> is in the span of the other two difference vectors in the determinant. You don't need a matrix or a determinant. The whole thing is just linear algebra.
  6. Jun 4, 2013 #5


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    Yes, the last column could be all 6's or w.e. You can't have a 4x3 because there is no determinant defined for that. The whole 4x4 thing is a pretty useless expression. It's just in there because it looks pretty. Use the 3x3 version and try to understand why it works. That's the important thing.
    Last edited: Jun 4, 2013
  7. Jun 5, 2013 #6
    Hello, So I have been experimenting and also have some more questions about the three by three.

    So if you look at the link above for the three by three matrix I say that it is the same as this form.

    I don't know how to do matrices so please excuse this.
    I will write it left to right but please know I mean row one, row two, row three....

    So I have.

    x,y,z ax,ay,az bx,by,bz = rx,ry,rz ax,ay,az bx,by,bz

    So I have a question though. I believe this right. I want to know though if we look at a vector equation of a plane. Say we have x = some general point on the plane and v(hat) is a normal unit vector we got from crossing two vectors parallel to our plane and doing the whole v/|v|. Oh and r is a point on the plane. So we have this

    x dot v(hat) = r dot v(hat)

    Now in this equation we use a unit vector but when I did a numerical example from determinate form I get the same answer as if I had used a unit vector in my determinate. I'm confused on this.
  8. Jun 5, 2013 #7


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    Just so you know, the word is "determinant".
  9. Jun 5, 2013 #8
    OK thanks do you have any advise for my last post?
  10. Jun 5, 2013 #9


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    I haven't been following this thread, but I think your question relates to what I told you before. If ##\vec X = \langle x,y,z\rangle## and ##\vec P = \langle a,b,c\rangle## are position vectors to points on a plane, and ##\vec N## is a normal vector to the plane, then the equation of the plane is ##(\vec X - \vec P)\cdot \vec N = 0##. (You really should learn that). It doesn't matter whether ##\vec N## is a unit vector or not for this calculation because if it isn't it just has the effect of multiplying the equation through by a constant.
  11. Jun 5, 2013 #10


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    It's not very difficult to write matrices using LaTeX. Click anywhere in the equation to see what I did.
    $$\begin{bmatrix} x & y & z \\ ax & ay & az\\ bx & by & bz\end{bmatrix} \
    = \begin{bmatrix} rx &r y &r z \\ ax & ay & az\\ bx & by & bz\end{bmatrix}$$
    It doesn't make sense to dot points with vectors. You have described x and r as points in the plane.

    Now if P and R are points in the plane, then ##\vec{OP}## and ##\vec{OR}## are vectors from the origin to P and R, respectively.

    If ##\vec{OP} \cdot \hat{v} = \vec{OR} \cdot \hat{v}##, then there's not a lot you can say about the two vectors.

    If both OP and OR are unit vectors, then all you can say is that both vectors make the same angle with v, or more precisely, the cosines of the angles are the same.
    I'm confused on what you're asking.
  12. Jun 5, 2013 #11
    OK, I'm just going to write out the whole question to see if I did it proper.

    The vectors a of norm a and b of norm b are parallel to a plane ∏ and R is a point on with position vector r = <rx,ry,rx> Find an equation in determinant form for a general point x = <x,y,z> of the plane involving the components <ax,ay,az> of a and <bx,by,bz> of b and <rx,ry,rx> of r. Hence find the distance,d ,of the plane from the origin and the linear equation of the plane in terms of the Cartesian components x,y,z of a general point and the components of a and b.

    So, First determinant form. I came up with

    \begin{bmatrix} x & y & z \\ ax & ay & az\\ bx & by & bz\end{bmatrix} \
    = \begin{bmatrix} rx &r y &r z \\ ax & ay & az\\ bx & by & bz\end{bmatrix}

    Edit: I tried to past the code from mark 44 post please just look at that because it is the form I had.

    Now to get d you have to consider the expansion of the right side of the equation above.
    First expand it then divide it by the norm of a cross b = v.

    Expanding the matrix I got ...
    rx(aybz - byaz) - ry(aybz - byaz) + rz(axby - bxay)

    Now the norm of |v| = √(aybz - byaz)^2 -(aybz - byaz)^2 + (axby - bxay)^2

    All together d = rx(aybz - byaz) - ry(aybz - byaz) + rz(axby - bxay) / √(aybz - byaz)^2 -(aybz - byaz)^2 + (axby - bxay)^2

    Lastly the linear equation of the plane . I just expanded the left side.

    (aybz-byaz)x -(axbz-bxaz) + (axby-bxay)z = rx(aybz - byaz) - ry(aybz - byaz) + rz(axby - bxay)

    Is the correct? Thanks
    Last edited: Jun 6, 2013
  13. Jun 6, 2013 #12
    Is the above correct? please and thanks
  14. Jun 6, 2013 #13


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    I'm really not interested in your determinant stuff. But regardless of that, I am really disappointed after what we have talked about in this and other threads that you even have to ask if that is the correct equation of a plane. Does it look like the equation of a plane?
  15. Jun 6, 2013 #14
    Looks like a plane to me. Also sir for you to express disappointment in me in unnecessary. You do not know my learning history. How are you certain that my lack of catching on is not due to a learning disability? I would like to remind you that my user name is Jbreezy and perhaps in the future you might be better off not to open my posts and respond if I'm such a thorn in your side.
    The determinate form was because that is what the question asked.
  16. Jun 7, 2013 #15

    Ray Vickson

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    The fact is that what you wrote is NOT the equation of a plane: you have nonlinear terms that (AS WRITTEN) involve products of the variables x, y and z, and so is a highly nonlinear surface. If you really meant ##a_x, a_y, a_z, ## etc., you should take the trouble to write them properly, so others can figure out what you mean (and what you know): writing ##ax, ay, az## means you are writing products. If you don't want to use TeX, use instead the "X2" button at the top of the input panel, or if you prefer, just write a_x. That way, everybody will know what you mean.

    BTW: the word is determinant, not determinate.
  17. Jun 8, 2013 #16
    I mean a_x, a_y, a_z so it isn't nonlinear then it is OK?
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