Lines in Space

Key Questions

  • What is the symmetric equation of a line in three-dimensional space?

    The symmetric equation of the line with the direction vector vec{v}=(a,b,c) passing through the point (x_0,y_0,z_0) is:

    {x-x_0}/a={y-y_0}/b={z-z_0}/c,

    where none of a,b and c are zero.

    If one of a,b, and c is zero; for example, c=0, then we can write:

    {x-x_0}/a={y-y_0}/b and z=z_0.

    If two pf a,b, and c are zero; for example, b=c=0, then we can write:

    y=y_0 and z=z_0

    (There is no restriction on x, it can be any real number. )

    I hope that this was helpful.

    Wataru · 1 · Nov 20 2014
  • What is meant by the parametric equations of a line in three-dimensional space?

    We want to find the parametric equations of the line L passing through the point P and parallel to a vector A.

    Let us take a 3-dimensional point in R^3, call it P = (x_0,y_0,z_0).

    A line L is drawn such that it passes through P and is parallel to the vector A = (u,v,w).

    3 parametric equations can be written which express the components:

    x = x_0 + tu
    y = y_0 + tv (-oo < t < oo)
    z = z_0 + tw

    As an example, with a point P = (2,4,6) and vector A = (1,3,5), we have the following parametric equations:

    x = 2 + tu
    y = 4 + 3v (-oo < t < oo)
    z = 6 + 5w

    Parametric Equation Calculator

    Don S. · · Aug 20 2014
  • How do I determine whether a line is in a given plane in three-dimensional space?

    Answer:

    For plane:

    ax+bx+cx+d = 0

    and line:

    (x, y, z) = (x_0, y_0, z_0) + t(u, v, w)

    these are suitable conditions:

    { (ax_0+by_0+cz_0+d = 0), (au+bv+cw = 0) :}

    Explanation:

    The most general equation of a plane in three dimensional space is:

    ax+by+cz+d = 0

    A line in three dimensional space can be represented in a variety of ways, but one representation that will work for any line is the parametric form:

    (x, y, z) = (x_0, y_0, z_0) + t(u, v, w)

    This line will lie in the plane if and only if two distinct points of it both satisfy the equation of the plane.

    Using t = 0 and t = 1, we obtain the conditions:

    { (ax_0+by_0+cz_0+d = 0), (a(x_0+u) + b(y_0 + v) + c(z_0 + w) + d = 0) :}

    Subtracting the first of these from the second, we obtain the condition:

    au+bv+cw = 0

    So a necessary and sufficient set of conditions is:

    { (ax_0+by_0+cz_0+d = 0), (au+bv+cw = 0) :}

    George C. · 1 · Sep 17 2015
  • How do I find the intersection of two lines in three-dimensional space?

    I'll show how to solve this problem using an example using the vector based form of a two straight lines. Bear in mind that there will be one of the following outcomes:

    • a single unique point.
    • no solution (if the lines do not intersect).
    • infinitely many solutions (if the lines coincide).

    Suppose we have;

    L_1: \ \ \ \ vec(r_1) = ((5),(2),(-1)) +lamda((1),(-2),(-3))

    L_2: \ \ \ \ vec(r_2) = ((2),(0),(4)) +mu((1),(2),(-1))

    It is vital that the variable parameters (lamda and mu) have different symbols, as they are different parameters relative to each line.

    If the lines do meet then for some specific values of lamda and mu then:

    vec(r_1) = vec(r_2)

    In which case we would have:

    ((5),(2),(-1)) +lamda((1),(-2),(-3)) = ((2),(0),(4)) +mu((1),(2),(-1))
    :. \ \ \ \ \ \ ((5+lamda),(2-2lamda),(-1-3lamda)) = ((2+mu),(2mu),(4-mu))

    By comparing the coefficients of hat(i), hat(j) and hat(k) we have three equations in two unknowns, so we can solve for lamda and mu using the first two equations and check if the third is satisfied:

    hat(i) : \ \ \ \ \ \ \ \ \ \ \ 5+lamda = 2+mu " " ....... [1]
    hat(j) : \ \ \ \ \ \ \ \ \ 2-2lamda = 2mu " " ....... [2]
    hat(k) : \ \ \ \ -1-3lamda = 4-mu \" " ....... [3]

    Eq [1] + 1/2Eq[2] gives => 6=2+2mu => mu=2
    Subs mu=2 into Eq[1] => 5+lamda=4=>lamda=-1
    Subs mu=2 and lamda=-1 into Eq [3]=>-1+3=2=4-2

    So we have established that if we choose lamda=-1 and mu=2 then we get a unique solution simultaneously satisfying all three equations.

    We can then substitute lamda=-1 into L_1, (equally mu=2 into L_2 would work) to determine the actual coordinate of intersection.

    vec(r_1) = ((5),(2),(-1)) -((1),(-2),(-3))
    \ \ \ = ((5-1),(2+2),(-1+3))
    \ \ \ = ((4),(4),(2))

    so in this example, the lines L_1 and L_2 intersect at the coordinate (4,4,2)

    Steve M · 1 · Jan 4 2017
  • How do I find the slope of a line through two points in three dimensions?

    Since we are talking about a line in 3-D, it is more appropriate to talk about its direction vector than its slope. To find the direction vector vec{v} of the line passing through two points (x_1,y_1z_1) and (x_2,y_2,z_2) can be found by

    vec{v}=(x_2-x_1,y_2-y_1,z_2-z_1).

    Wataru · 1 · Sep 26 2014

Questions

3-D Cartesian Coordinate System