Equation of a Line in 3-D SpaceDate: 03/18/2004 at 00:24:42 From: Jared Subject: 3-D Graphing Equation (x,y,z) Given two points in 3-D space, such as A(x1,y1,z1) and B(x2,y2,z2), what would be the equation of the line that connects those points? I know that in the 2-D plane the equation of a line in slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. Is there something similar in 3-D? I believe the equation may involve finding the intercepts of all three planes (xy, yz, and xz). Date: 03/18/2004 at 12:28:34 From: Doctor Peterson Subject: Re: 3-D Graphing Equation (x,y,z) Hi, Jared. In two dimensions, the equation y = mx + b defines a line; but vertical lines can't be represented that way. The most general form of a line is Ax + By = C Here A and B together define the direction of the line. If you've heard of vectors, the vector (A,B) is perpendicular to the line; otherwise you can just say that the line from (0,0) to (A,B) is perpendicular to it. C determines where the line crosses that perpendicular line. The equivalent of the first form in three dimensions is z = mx + ny + c with a sort of slope for both m and n, and an "xy-intercept" c. (The terminology is not standard, because we don't commonly use this form.) Note that this is NOT a line, but a plane. The general form for a plane is Ax + By + Cz = D Here the vector (A,B,C) is perpendicular to the plane. To define a line, you need to intersect two planes. Does that make sense? So one way to define a line is with TWO equations. You might use these, just as an example: z = mx + b z = ny + c The first is a plane parallel to the y axis, and the second is a plane parallel to the x axis. Not all lines, of course, can be written with this form. You can see some of the standard forms for a line here: http://mathforum.org/dr.math/faq/formulas/faq.ag3.html#threelines One of those is the two-point form, which answers your question: (x - x1)/(x2 - x1) = (y - y1)/(y2 - y1) = (z - z1)/(z2 - z1) Note that there are actually two equations here (equating three expressions). This is a special case of the point-direction form (x - x1)/a = (y - y1)/b = (z - z1)/c This equation is best described in terms of vectors. The vector (a,b,c) points in the direction of the line; the equations say that the change in each coordinate is proportional to the coordinates of this vector. If you introduce a new variable t (which you can think of as time while you travel along the line), and let that be equal to the three equal expressions above, (x - x1)/a = (y - y1)/b = (z - z1)/c = t then you can solve for x, y, and z to get the parametric equations of the line (expressing x, y, and z in terms of the parameter t): x = at + x1 y = bt + y1 z = ct + z1 This is probably the most-used form. You can see that it says that you can find the coordinates at time t by multiplying a, b, and c (the components of velocity) by the time, and adding the coordinates of the starting point. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/