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Parameter Within A Linear Equation Conjecture

  1. Aug 29, 2006 #1


    X2 + Y2 + Z2 + D = 0

    WHERE t = parameter = -D

    V = direction vector


    Vt = <at, bt, ct>

    It seems as if it is...but I can't seem to prove it.
    Last edited: Aug 29, 2006
  2. jcsd
  3. Aug 29, 2006 #2


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    It's not at all clear to me what you are trying to say.
    What are X2, Y2, Z2? Since you say this is a linear equation, you surely don't mean X2, etc. Do you mean X2, Y2, Z2? If so I don't see any need for the subscripts. Why not just x+ y+ z+ D= 0?

    Do you mean allowing D to change? The equation, x+ y+ z+ D= 0 or, equivalently, x+ y+ z= -D, is the equation of a plane. Different values of D give different planes all parallel to one another. In that situation, yes, D (or -D) is a parameter.

    What direction vector? A normal vector to the planes above?
    Yes, if <a, b, c> is a vector then <at, bt, ct>, for different values of t, gives vectors of different lengths along the same line. But what does that have to do with -D above?
  4. Aug 30, 2006 #3
    to describe a plane, one uses two position vectors to locate two points on the plane.

    r = ro + vt where vt = direction vector and vt is the line that connects the two points at the ends of the position vectors. also, t = parameter that is the scalar multiple of the original direction vector.

    vt is the line on the plane we are trying to describe, and vt can be described in terms of r and r0. vt = r - ro
    do the dot product between the normal vector and vt and you get the parametric equations.

    the parametric equations lead to the linear equation.
    if x0, y0, z0 = 0...then,
    x2 + y2 + z2 = t

    or, at least i believe it does.

  5. Aug 30, 2006 #4


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    Not quite. Two points do not determine a plane. Three points are required. You may be thinking of using two vectors from one point in the plane to two other points. Those are not "position vectors" (position vectors are vectors from the origin of the coordinate system to the points in question).

    Okay, so this is the vector between two points in the plane (again, NOT a "position vector"). But then two points only give you one vector.

    This is the first time you've mentioned the normal vector. If you are given THAT and a single point in the plane, that's enough to describe the plane.

    It's still not clear to me what "x2, y2, z2" represent- I asked that before.
    And WHAT "parametric equations" are you talking about?
    Also the dot product between the normal vector to the plane and any vector in the plane must be 0: the dot product of any two perpendicular vectors is 0. I don't see where that "t" came from.

    Not until you explain what "x2, y2, z2" mean!

    If the point [itex](x_0, y_0, z_0)[/itex] is in the plane and (x,y,z) is any point in the plane, then the vector between them, [itex](x-x_0)i+ (y- y_0)j+ (z- z_0)k[/itex] is in the plane and so is perpendicular to Ai+ Bj+ Ck. Since they are perpendicular, their dot product is 0:
    [tex](x-x_0)i+ (y- y_0)j+ (z- z_0)k)\cdot(Ai+ Bj+ Ck)= A(x-x_0)+ B(y-y_0)+ C(z- z_0)= 0[/tex]
    Of course, we can multiply that out and get the equation
    [tex]Ax+ By+ Cz= Ax_0+ By_0+ Cz_0= D[/tex]
    where D is defined as [itex]Ax_0+ By_0+ Cz_0[/itex].

    That has nothing to do with "parametric equations".
    Since a plane is 2 dimensional, a plane in a 3 dimensional coordinate system can be written in 3-2= 1 equation using just the coordinates x,y,z while, since it is 2 dimensional would require 2 parameters.

    A single line, which is 1 dimensional would require 3-1= 2 equations in x,y,z (we could think of the two equations as describing two planes which intersect along the given line) but requires only 1 parameter.
  6. Aug 30, 2006 #5
    x2 means x squared

    first of all, I'm sorry to have yelled at you...didn't mean it....especially to a professor.

    to describe a plane, one uses two position vectors to locate two points on the plane. that line that forms on the plane must be used to form the dot product with the normal vector of that line. yes...it will be zero.

    here are the position vectors.

    ro = <xo, yo, zo>
    r = <x, y, z>

    the line between the ends of these two points form on the plane we want to describe. let's call that vector p. let's say that p has the same components as the direction vector, v.
    v = <a, b, c> = p
    p = <at, bt, ct> and in this case...t =1 because we said that p = v.

    ro + p = r
    ro + v = r

    <xo + at, yo +bt, zo +ct> = <x, y, z>

    these are your parametric equations.

    because we said that v = p
    <xo + a, yo + b, zo + c> = <x, y, z>

    now...we do the dot product to find the LINEAR EQUATION OF THE PLANE.

    p = r - ro

    n = normal vector = <h, i, j>

    n dot p = 0

    n dot (r - ro) = 0

    that becomes hx + iy + jz = h(xo) + i(yo) + j(zo)
    which simplifies to

    h(x - xo) + i(y - yo) + j(z - zo) = 0

    this is where i am stuck....i'm not very sure what happened to D. if you find out what happened to D...is it the parameter???

    thanks for the help.
    Last edited: Aug 30, 2006
  7. Aug 31, 2006 #6


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    Yelling? I didn't notice any yelling- I thought you were being very polite! You should see me when I yell!

    Still not clear what you mean. Two points and two position vectors do not determine a plane. One line in the plane is not enough to determine a plane (imagine rotating the plane around the given line). A single line has an infinite number of (unit) normal vector. There is a plane normal to the line and any vector in that plane is a normal vector to the line.

    Yes, those are parametric equations for the line between your two points (strictly speaking the "parametric equations" are x= x0+ at, y= y0+ bt, z= z0+ ct. What you give is a vector equation for the line) NOT equations for a plane which is what I thought you were saying.

    Okay, now I see what you are doing. However, you are not just using just two points in the plane as you said before. You are assuming that you are given two points in the plane and the normal vector to the plane.

    Actually, as I said before, if you are given the normal vector, the one point is sufficient. The "vector" you need in the plane is between the given point [itex]x_0i+ y_0j+ z_0k[/itex] and the "general point" xi+ yj+ zk where x, y, z are variables. Then your "r" is [itex]r= (x-x_0)i+ (y-y_0)j+ (z-z_0)k. With normal vector Ai+ Bj+ Ck.

    (you are confusing the issue by writing the normal vector as "<h, i, j>= hx+ jy+ kz" where you seem to be using x, y, z as basis vectors. That is not standard notation and if you do that, you cannot then use x, y, z as variables. Standard notation is the x, y, z represent coordinates of a variable point in the plane and the unit vectors in the x, y, z directions are i, j, k respectively.)

    Writing the normal vector as <A, B, C>= Ai+ Bj+ Ck, the dot product becomes [itex]<A, B, C> \cdot <x-x_0, y-y_0,z-z_0>[/itex] or, more formally [itex](Ai+ Bj+ Ck)\cdot (x-x_0)i+ (y-y_0)j+ (z-z_0)k[/itex] both of which equal A(x-x_0)+ B(y-y_0)+ C(z-z_0). Since the two vectors are perpendicular, the product is 0: [itex]A(x-x_0)+ B(y-y_0)+ C(z-z_0)= 0[itex].
    Multiplying that out gives [itex]Ax+ By+ Cz- Ax_0- By_0- Cz_0= 0[/itex] Which we can write as [itex]Ax+ By+ Cz= Ax_0+ By_0+ Cz_0[/itex]. Since A, B, C, x0, y0, z0 are simply numbers (as opposed to the variables x, y, z) Ax0+ By0+ Cz0 is simply a number. Call it "D" and you get Ax+ By+ Cz= D. No, D is not a parameter- it is a fixed numbers and does not vary. The equation Ax+ By+ Cz= D is not a parametric equation.

    Example: write the equation for the plane having normal vector 3i+ 2j+ 4z= <3, 2, 4> and containing the point (1, 1, 3). If (x, y, z) is a variable point in the plane, then (x-1)i+ (y-1)j+ (z-3)k= <x-1, y-1, z-1> is a vector in the plane. The dot product with the normal vector is
    [itex](3i+ 2j+ 4k)\cdot((x-1)i+ (y-1)j+ (z-3)k)= 3(x-1)+ 2(y-1)+ 4(z-3)= 0[/itex]. That, of course, is the same as 3x- 3+ 2y- 1+ 4z- 3= 3x+ 2y+ 4z -(3+ 1+ 3)= 0 or 3x+ 2y+ 4z= 7. "D" is just the number 7 and is not a parameter. This is not a "parametric" equation.

    You can write this plane in terms of parametric equations but since a plane is 2 dimensional, you need two parameters. One easy way (there are always an infinite number of parametric equations for a surface depending on choice of parameter) is to take x and y themselves as parameter. That is, write x= u, y= v and the the equation above for the plane becomes 3u+ 2v+ 4z= 7 so 4z= 7- 3u- 2v and the parametric equations are x= u, y= v, z= (7/4)- (3/4)u- (1/2)v.

    "Parametric equations" are equations of the form "x= f(u,v), y= g(u,v), z= h(u,v)"- a different equation for each coordinate, depending (in the two dimensional case) on two parameters (parametric equations of a line or curve depend on one parameter). An equation of the form Ax+ By+ Cz= D is not a "parametric equation" and has no parameter.
  8. Aug 31, 2006 #7
    then, what can we say about 7...or D besides it's merely being a number? isn't there any relationship between D and t?

    D = xo + yo + zo is there any meaning to this????? perhaps...a shift of the direction vector? I just appreciate the relationship between the direction vector and the linear equation very much...
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