What is the equation for a line in 3D space?

The line goes from (5, 5, 5) to (-8, -6, -3).

5 Answers

  • 1 decade ago
    Best Answer

    I'll label your points A(5, 5, 5) and B(-8, -6, -3). Now, we need a direction vector, which will be AB (you can use BA if you want but I'll use AB):

    AB = [-8-5 , -6-5 , -3-5]

    AB = [-13, -11, -8]

    Now that we have this, we'll use point A to find the equation of the line. I'm going to be finding the equation in parametric form (you can convert it to cartesian or vector later on if you wish):

    The equation is given be:

    x = x1 + at

    y = y1 + bt

    z = z1 + ct

    Where the line goes through (x1, y1, z1) and has direction vector [a, b, c]

    So, using your point (5, 5, 5) and direction vector [-13, -11, -8]:

    x = 5 - 13t

    y = 5 - 11t

    z = 5 - 8t

    Where t is the parameter and can take any real values.

  • 4 years ago

    This Site Might Help You.


    What is the equation for a line in 3D space?

    The line goes from (5, 5, 5) to (-8, -6, -3).

    Source(s): equation line 3d space: https://tinyurl.im/cH4w9
  • kullas
    Lv 4
    3 years ago

    Equation Of Line

  • 1 decade ago

    To describe a line in space we use vectors and parametric equations:




    where <a,b,c> represents the vector parallel to the line through the point (x,y,z)

    in your case let's say P=(5,5,5) and Q=(-8,-6,-3)

    Then vector PQ= (-8-5,-6-5,-3-5) = <-13,-11,-8>

    Using vector and point P we have the equations:




    For t=0 we get the point P; and t=1 gives point Q

    Source(s): Shenk Calculus and Analytic Geometry
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  • Anonymous
    6 years ago

    All the answers bellow are incorrect. For the parametric equation of a line, you need a direction vector, which is the normal to the plane in which the line is located. Those answers are wrong!

    • 5 years agoReport

      You're thinking of the equation of a plane. For a line the vector parallel to the line and a point on the line is needed for the parametric equations.

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