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Homework Help: Finding the parametric equation of a plane

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the equation of the following planes in cartesian and (vector) parametric form:
    a) the plane through the point (1,4,5) and perpendicular to the vector (7,1,4)
    b) the plane through the origin and the points (1,1,1) and (1,2,3)

    3. The attempt at a solution

    For part a)
    I found the cartesian form of the equation using a(x-x0) + b(y-y0) + c(z-z0), to give:
    7(x-1) + 1(y-4) + 4(z-5) = 0, which expands to:
    7x + y + 4z = 31, which is correct.

    However, I've no idea how to find the parametric form. The answer is given as r = (1,4,5) + s(4,0,-7) + t(0,4-1).

    Any help on how I'd go about answering this question?

    For part b)
    To find the cartesian form, I found the vectors from the origin to (1, 1, 1,) and (1, 2, 3), using V1->V2 = V2 - V1, giving me V0->V1= (1,1,1) and V0->V2 = (1,2,3). Using the cross product to get the orthogonal, then dotting that with (0,0,0) gives the correct equation, x-2y+z=0, which is correct.

    Again, I've got no idea how to find the parametric equation.

    Any help would be greatly appreciated. These aren't assignment questions or anything like that - just revision for my L.A exam next week.
  2. jcsd
  3. Feb 10, 2009 #2
    Let's pluck a plane out of the air: x + 3y - 7z = 20.

    To find the parametric equations for x, y and z is just to solve the equation as you would simultaneous equations, remember that?

    One equation, three unknowns => 3-1=2 parameters. Let z=t, y = s, so x = 20 - 3y + 7z = 20 - 3s + 7t. Hence (x, y, z) = (20 - 3s + 7t, s, t) = (20, 0, 0) + (-3, 1, 0)s + (7, 0 , 1)t.
  4. Feb 10, 2009 #3


    Staff: Mentor

    For a, for the parametric form you want, you have one point in the plane, Q(1, 4, 5). You need to find two additional points in the plane, call them R and S.

    The points R and S should be such that QR is not parallel to QS.

    Then any point in the plane satisfies P = OQ + r*QR + s*QS.

    Here's what's going on geometrically. To get from the origin to any point in the plane, I have to get to some point on the plane (OQ does that), and from there, some linear combination of QR and QS gets me to the exact point. The parameters of this parametric form are r and s.
  5. Feb 10, 2009 #4
    Applying this method to a), I get (x,y,z)=(31-4t-7s,s,t)= (31,0,0)+(-7,1,0)+(-4,0,1), which differs from the answer given (shown in the OP). How would we answer the question without having to use a different position (as seems to be the expectation in all the questions)?
  6. Feb 10, 2009 #5
    OK, I get that much. But how would be find the points QR and QS?

    I'm guessing one possible method would be to use the dot product, such that QR . (7,1,4) and QS . (7,1,4) = 0. Is there an easier approach to the question (and one that doesn't give me infinite possible solutions, which I'm guessing would make life pretty hard for the examiner:devil:)?

    Thank you both for your help.
  7. Feb 10, 2009 #6


    User Avatar
    Science Advisor

    Your answer is incorrect because taking s= t= 0 gives (31, 0, 0) which does NOT satisfy 7x- y+ 4z= 31. Did you forget to divide by 7? Solving the cartesian equation for x we have x= (y- 4z+ 31)/7 and, taking y and z as the parameters, y= s, z= t, x= (s- 4t+ 31)/7 so
    r= ((s- 4t+ 31)/7, s, t)= (31/7, 0, 0)+ (1/7, 1, 0)s+ (-4/7, 0, 1)t.

    There exist an infinite number of parametric equations for anything because the choice of parameter is arbitrary. I, because I dislike fractions, might have chosen to solve for y rather than x: y= x+ 4z- 31 so taking x= s, z= t, r= (s, s+ 4t- 31, t)= (0, -31, 0)+ (1, 1, 0)s+ (0, 4, 1)t.

    In the answer your book gives, r = (1,4,5) + s(4,0,-7) + t(0,4,-1), they have not solved the cartesian equation but worked directly from the given information. Any point in the plane can be written as a single point in the plane (and (1, 4, 5) is given) plus a linear combination of two independent vectors in the plane. Any vector on the plane must be perpendicular to (7, 1, 4) and we can be sure two vectors are independent by taking one (1, 0, a) and the other (0, 1, b). For what a is (1, 0, a) perpendicular to (7, 1, 4)? We must have 7+ 4a= 0 or a= -7/4. (1, 0, -7/4) will do, or since I dislike fractions, (4, 0, -7). A vector of the form (0, 1, b)is perpendicular to (7, 1, 4) if 1+ 4b= 0 so b= -1/4. Such a vector is (0, 1, -1/4) as well as (0, 4, -1). We can write any point on the plane as the intial point, (1, 4, 5) plus a linear combination of those two vectors in the plane: (1, 4, 5)+ (4, 0, -7)s+ (0, 4, -1)t.
  8. Feb 10, 2009 #7
    Thanks heaps for clearing that up, HoI. On your first paragraph, I did solve for y and let x and z equal arbitrary constants, but I apparently wrote the order as (y,x,z) instead of (x,y,z). :redface:
  9. Feb 11, 2009 #8
    Alright - one more question.

    If we're given a point and a vector parallel to the plane, how would we find the vector equation?

    The question is: Find the vector equation of the plane through the point (6, 5, -2) and parallel to the plane x + y - z + 1 = 0.

    How I went about solving it was find a vector perpendicular to the plane using the dot product:
    (a,b,c)=(1,0,1) satisfies the above, thus it is perpendicular to the plane.

    Finding two vectors perpendicular to this (or one, as we already have a parallel vector) would allow us to make a linear combination along the plane and thus define the plane.
    Therefore, (a,b,c)=(1,0,-1), with only the one possible independent vector, using the parallel vector given in the question:
    (6,5,-2) + s(1,0,-1) + t(1,1,-1) (equation of the plane)

    However, the answer given is:
    (6,5,-2) + s(1,0,1) + t(0,1,1)
    Now both the vectors used in the equation are perpendicular to the plane, so how is it possible that the above is an equation for the plane? Unless I'm missing something, I'm guessing it's just an error on their behalf?
  10. Feb 11, 2009 #9


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    If two planes are parallel, they have the same normal vector. Find the plane containing (6, 5, -2) with normal vector <1, 1, -1>. You seem to be forgetting that for a plane written as Ax+ By+ Cz= constant, <A, B, C> is the normal vector.

    (x- 6)+ (y- 5)- (z+ 2)= 0 so x+ y- z= 13 is the "Cartesian" equation. Solving form z, z= x+ y- 13 so x= s, y= t, z= s+ t- 13 are parametric equations for the plane and <0, 0, -13>+ <1, 0, 1>s+ <0, 1, 1>t is a vector equation.

    No. (1, 1, -1) itself is perpendicular to the given plane.

    Again, (1, 1, -1) is perpendicular to the given plane so (x, y, z) is parallel to the plane if and only if x+ y- z= 0. z= x+ y in that case. Taking x= 1, y= 0, z= 1 so (1, 0, 1) is parallel to the plane. Taking x= 0, y= 1, z= 1 so (0, 1, 1) is parallel to the plane. Since (6, 5, -2) is a point in the plane, (6,5, -2)+ s(1, 0, 1) is in the plane for all s and (6, 5, -2)+ t(0,1,1) is in the plane: since the plane is two dimensional all points in the plane are of the form (6, 5, -2)+ s(1, 0, 1)+ t(0, 1, 1) for all s and t.
    Last edited by a moderator: Feb 11, 2009
  11. Feb 11, 2009 #10
    You really are a legend HoI. The section is barely covered in the textbook and we skimmed over it in the lectures. Thanks heaps for your help. *insert thumbs up smilie here*
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