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Does there exist a pair of interesecting lines that arent coplanar

  1. Oct 3, 2012 #1
    Suppose you have two lines that intersect at the point O. Let's say the lines are OP and OQ. Could you construct a plane that contains both these lines?

    I can visually imagine this happening in the 3-space, but I may be wrong. I was just wondering, because this allows one to define an angle between two lines in 3-space if I remember correctly.

    What about in an n-space? Must two lines form a plane in an n-space?

    I imagine that in the same way two points define a line in n-space, two lines can define a plane in n-space, but because of my limited knowledge of linear algebra, I can't prove this.

    But I'm hoping to learn, perhaps someone can offer me their insights.

  2. jcsd
  3. Oct 4, 2012 #2
    Yes, this is indeed true. Let's prove it. But to prove that, we need to know what a line is. What would you think the definition of a line is in n-dimensional space? Let's work with lines through the origin since it's easier.
  4. Oct 8, 2012 #3
    Have you given up on this?

    For n dimensions a straight line is represented by a set of (n-1) simultaneous equations.

    It is only represented by a single equation in 2D.

    In general

    [tex]\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}} = ...........[/tex]

    Which is complete for n=3 (2 equations) and continued by the dots.
  5. Oct 10, 2012 #4


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    Another way of looking at it, in three space, is this: given two intersecting lines we can define a vector in the direction of each line. The cross product of the two vectors is perpendicular to both and the plane defined by that perpendicular vector and the point of intersection of the two lines contains both lines.

    For example, we can always set up a coordinate system so that the origin is at the point of intersection and the x-axis extends along one of the lines. Parametric coordinates for that line, in that coordinate system, are x= t, y= 0, z= 0. We can take x= as, y= bs, z= ds as parametric equations for the other line. <1, 0, 0> is a vector in the direction of the first vector and <a, b, c> is a vector in the direction of the other line. Their cross product, <0, -c, b> is perpendicular to both and the plane -cy+ bz= 0 contains both lines.

    As for defining an angle in n-space, most commonly we define it by exending the formula for dot product in two and three dimensional spaces:
    [itex]u\cdot v= |u||v|cos(\theta)[/itex] where |u| and |v| are the lengths of vectors u and v and [itex]\theta[/itex] is the angle between them. As I say, we can prove that is true in two and three dimensions and then use it as definition of "angle" between two lines in higher dimensions.
  6. Oct 10, 2012 #5
    Thank you for that comprehensive reply! Therefore, am I correct in concluding that for any two vectors there exsits a plane on which they both lie?

    If one or more of the vectors is a zero vector, then any plane containing the other vector must necessarily contain the first vector.

    The vectors, whether parallel or not, must intersect because they can translated to have their terminal points at the origin. If the vectors intersect, the terminal points of each vector along with the point of intersection form a triangle. Since a triangle lies on a plane, the vectors must be coplanar.

    If the vectors are identical, then any plane containing a line which in turn contains either vector must contain both vectors. We know that there exists a plane containing any line in n-space and that there exists a line containing any vector in n-space. (is this true?)

    Is this reasoning correct? In n-space also?

  7. Oct 10, 2012 #6
    I wish HOI hadn't introduced vectors, I find they confuse things.

    Any single line in space sweeps out a plane.

    A plane divides n (n>2) space into two halves.

    Consequently if you have two lines the planes they sweep out either intersect or are parallel.

    Two planes intersect in a line.

    Two lines may neither intersect nor be parallel in any space of dimension greater than or equal to 3.
    These are called skew lines.
    Last edited: Oct 10, 2012
  8. Oct 10, 2012 #7


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    Following on from what Studiot said, you can find these intersections using normal matrix row-reduction and you can determine skew-lines by seeing if you get an inconsistent solution to your set of equations.
  9. Oct 11, 2012 #8


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    It is a fundamental theorem of three dimensional Euclidean geometry that two intersecting lines lie in a unique plane. It is also true that two parallel lines lie in a unique plane. Two lines that are not parallel and do not intersect, "skew" lines, do not lie in a plane.
  10. Dec 3, 2012 #9
    Is this really true? So am I correct in guessing one could prove that the amount of space on one side of a plane is equally as infinite as on the other side??
  11. Dec 3, 2012 #10
    A point divides a line into two parts.
    Each part goes from the point to 'infinity'

    Formally we call these parts intervals.
    The 'infinite' end is open.
    We can include the whole line by including the point in one interval or other, but not both.
  12. Dec 3, 2012 #11
    right, that is pretty intuitive isn't it. Could a line segment divide a number line into uneven parts? Considering the line segment is finite
  13. Dec 7, 2012 #12


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    As I remember it is actually a postulate of Euclidean geometry that two intersecting lines determine a plane, not a theorem.

    The Cartesian coordinate system is set up to be a model of Euclid's postulates. One can see this from distance measurement along Cartesian straight lines because it obeys the Pythagorean theorem. Once you know this all else follows.
  14. Dec 12, 2012 #13


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    That depends upon what you mean by "plane". Some texts use the term "plane" to mean a "flat" subset of [itex]R^n[/itex] of dimension 2 no matter what n is and use the term "hyper-plane" to mean a "flat" subset of dimension n-1. Other texts use the term "plane" to mean a "flat" subspace of dimension n-1.
  15. Dec 12, 2012 #14
    Thank you Halls of Ivy for tightening up on my terminology I really should have said hyperplane, dimension (n-1), and reserved 'plane' for dimension n=2.
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