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