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How do you explain x+y+z=5 is a plane?

  1. Jul 28, 2006 #1
    I teach first year uni students and they don't seem to know what is a plane
    and what is a line in 3d. I have been paid to make an online multiple choice
    a prototype is given below. Any help would be appreciated

    (a) has infinite solutions
    examples of solutions are (-1,6), (0,5) , (1,4),(2,3) etc
    (b) has a unique solution

    2) x+y=0 could be represented by the parametric equation
    (a) (0,5) + (-1,1)t
    where t can take any value
    (b) (0,5) + (1,1)t
    where t can take any value

    3) x+y =5 can be represented by
    (1 1 |5)

    4) x+y=5
    (a) is a line
    (b) is a plane

    5) x+y+z=5
    (a) Has infinite solutions
    examples of solutions are (-1,0,6), (0,-1,6), (0,0,5),(1,0,4)
    (b) Has a unique solution

    6) x+y+z=5 could be represented by the parametric equation
    (a) (5,0,0)+(-1,1,0)s + (-1,0,1)t
  2. jcsd
  3. Jul 28, 2006 #2


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    My, my.. let me guess. You're actually a student, who needs help with these problems. Rather than explaining that, you're attempting to lie and tell us all you're a professor, and thus need our help answering your own questions... (?)

    Just for the record, we don't do homework assignment for you. On the other hand, if you explain your thoughts and let us know why you're stuck, we can help you become unstuck.

    - Warren
  4. Jul 28, 2006 #3
    Ask me hard question so I can prove my credentials please.
  5. Jul 28, 2006 #4
    OK I will explain the questions to prove I am not a liar

    can be represented in RREF as (1 1 |5)
    y is a non leading varible ( a free variable)
    let y =t
    so x+ t = 5
    to summarize)
    (x,y) = (5-t , t)
    this is the same as
    (x,y) = (5,0) + (-1,1)t
    This a line in R2

    Moving on to x+y+z=5
    This can be written as
    (1 1 1|5)
    this is in row echelon form
    y and z are non leading
    let y =s and z= t
    so x + s +t =5
    to summarize
    (x,y,z) = (5- s -t , s, t)
    which is the same as
    (x,y,z) = (5,0,0) + (-1,0,1)s + (0,-1,1)t
    this is a plane in R3
    it is ismorphic to R2 since (x,y) -> (5,0,0)+ (-1,1,0)x+(0,-1,1)y
    is one to one and onto.
    Last edited: Jul 29, 2006
  6. Jul 29, 2006 #5


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    Do you think it would help to start by showing that x=0, y=0 and z=0 represent planes through the origin? Then if your students are familiar with vector spaces, it's easy to contruct a linear combination and take it from there. But even if they're not, the progression becomes easier to make from that starting point (or in that case, from x=a, y=b and z=c).

    Alternatively you could just take the general equation of the plane and fix z at z=z1, say. Plug this into the equation and you are left with the equation of a line in 2d. This is easy to demonstrate pictorially or with props (that the intersection of any 2 planes in general, but also a general plane with a particular plane in this case, gives a line).
  7. Jul 29, 2006 #6


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    This might be a pedestrian suggestion, but why not just show them by drawing a graph? I suppose I've given basically the same suggestion as Gokul, but it may as well be stated twice. It's pretty easy to see when you just show a graph of the equation to see what it represents and why it represents it. I learned the shapes represented by different forms of equations all the way back in fourth grade by plotting them on a graph.

    To your test, showing some of the solutions below answer 1(a) kind of gives it away.
    Last edited: Jul 30, 2006
  8. Jul 29, 2006 #7
    Does anyone have sample wrong answers for the multiple choice test I was writing in the first post?

    I offer a warning to people that will teach that subject. The that bottom half of the class does not believe you initially when you say x=0 is a plane in R3. The idea that y and z can take any value goes against their usual high school experience of there being only one answer. It is beneficial to discuss that x=0 is a line in R2.
    Last edited: Jul 29, 2006
  9. Jul 29, 2006 #8


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    Then perhaps it might be a good idea to spend some time discussing various vector spaces of different dimensions, and how to identify them, paricularly looking at lines and planes as Euclidean 1- and 2-spaces (perhaps without actually getting into what makes a vector space, in gory detail).
    To what? I'm not sure I understand the source of the problem here. Perhaps it is in not completely appreciating the importance of the domain of a function?
    You mean a line, don't you? (Or did you mean R^3?)
    Last edited: Jul 29, 2006
  10. Jul 29, 2006 #9


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    Just to make sure that typo doesn't carry through into the final homework set, that should be (x,y) = (5,0) + (-1,1)t, no?

    Looks like a similar error here (or am I getting something wrong?) Shouldn't that be (x,y,z) = (5,0,0) + (-1,1,0)s + (-1,0,1)t?
  11. Jul 29, 2006 #10
    Gokul I am embarassed at my typos, I will fix them

    As for the dilema of students being used to only have one answer, If you
    if you are in R3 and say x=0 you can ask the students what values y and z can take and they don't know, when you say y and z can take any value they don't believe you initially, I assume this is because In high school there is never a question where the answer is both variables can take any value, they either get a point or a line.
  12. Jul 29, 2006 #11

    matt grime

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    It is just a matter of explaining what the words mean. Properly you are describing the set of points {(x,y,z) : x=0}. If they don't understand that this means y and z can take any value then that is because no one has told them what that set notation means. So tell them.
  13. Jul 30, 2006 #12


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    If you have any power to do so, you might also consider upping the pre-reqs for the class you teach if the students do not grasp elementary concepts at the start of the class.
  14. Jul 31, 2006 #13


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    to show some set is plane, it helps to know what you mean by a "plane". so you might begin by discussing what a plane is. how you recognize something is a plane. once they agree on some criterion that characterizes a plane, then check it on your set.

    for example, maybe they like parametric representations and will agree that if v is a vector, p is a point, and t is a number, then the set of points of foirm p + tv, for various numbers t, forms a line. Oops, even here we must say v is not the zero vector.

    similarly the set of points of form p + tv + sw, where s,t are numbers and v,w, are non zero vectors may be a plane or may be a line. why?

    anyway then one can show that the points (x,y,z) such that x+y+z = 5, are exactly those of the form (5,0,0) + t(1,0,-1) + s(0,1,-1), where s,t, are numbers. e.g. if x+y+z = 5, then Z = 5-X-Y, so
    (x,y,z) - (5,0,0) = (x-5,y,z) = (x-5).(1,0,-1) +y(0,1,-1) = (x-5,y,5-x-y)
    = (x-5,y,z), so indeed (x,y,z) = (5,0,0) + t(1,0,-1) + s(0,1,-1), where t = x-5 and s = y.

    vice versa, if (x,y,z) = (5,0,0) + t(1,0,-1) + s(0,1,-1) = (5+t,s,-s-t) then clearly the coordinates add up to 5.
  15. Aug 27, 2006 #14
    to describe a plane from a linear equation, you need to indicate two elements. first, you need the position vector that is perpendicular to the area of the plane, also known as the Normal vector. then, you need the location of the point in the plane.

    x + y + z - 5 = 0

    N = <1, 1, 1>
    P (0, 0, 5)

    that ought to be sufficient to indicate that the linear equation is a plane.
    Last edited: Aug 27, 2006
  16. Aug 27, 2006 #15


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    The fact that, given any two points, say (a, b, c) and (d, e, f) satisfying the conditions for the set, the entire line ((d-a)t+ a, (e-b)t+ b, (f-c)t+c), for all real t, also satisfies the conditions for the set shows at all lines through two points on the set is in the set- that's a plane.
  17. Sep 1, 2006 #16
    I must say I'm ashamed at some of the responses of this post..

    you say first year uni students...how about telling us what class this for first (I didn't know linear algebra was supposed to be first year)? I doubt that the students need a pre-req of knowing what a linear combination is when they are barely taking elementary, intermediate or even college algebra for that matter.

    Plus, the comment was made to change the pre-reqs for first year university curriculum....hahahahahahaha....

    Usually students are exposed to the xy-plane, thus you can build upon that..make a analogy to a paper...floor..whatever..i've seen teachers use the corner of a classroom as their xyz axis...

    also, you might want to show the kids a horizontal line...like y = 4.
    say that in the form y = mx + b, that b is 4, which is where the line is going to intersect the y-axis and that x is allowed to run...so that y = (0)(x) + 4...they can understand that if x is anything it will be multiplied by 0 which just makes y = 4 for any value of x then y = 4 and that's how the line is created...also tell them that when the variable is not mentioned it's implied that it can be any value, etc, etc....if your going to be an educator than perhaps you can stop going around trying to prove that your worthy of your title and just be worthy of your title....
    Last edited: Sep 1, 2006
  18. Sep 1, 2006 #17
    What university allows students to take linear algebra in their first year? I am required to finish three semesters of Calculus and Analytical Geometry and then take Differential Equations before I can take Linear Algebra.
  19. Sep 1, 2006 #18
    Every australian university I know basically forces most students to deal with the topics in this thread in first year.
  20. Sep 2, 2006 #19
    I kinda liked the following idea when I was learning to visualise planes/3d surfaces (disclaimer: what I say has been said before, in slightly different words, I think).

    Say we have some equation: x + y = 5. We will only draw dots on our graph paper representing [itex]\mathbb{R}^2[/itex] if the numbers x and y on that point satisfy our equation.

    Now we have: x + y + z = 5. So we will only draw dots in our space representing [itex]\mathbb{R}^3[/itex] if the numbers x, y and z satisfy our equation. A way to see this is say we choose x = 0; then we get y + z = 5, and so we have a simple 2D equation for that plane. Then choose x = 1, and we have y + z = 4, and another equation. This can be repeated, obviously, for other values and variables.
  21. Sep 2, 2006 #20
    I'm sure plenty of people learn linear algebra at first year uni. In UK, it's in the A-Level (age 16-17) Maths course.
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