3D GeometryDate: 11/17/97 at 10:10:54 From: Tim Fallon Subject: 3d geometry Given two lines in 3space, you can draw a line of minimum distance between the two. This line will be perpendicular to both lines. I know how to get the distance and direction of this line, but I want to locate the line in 3space so that I can find its midpoint. Any help you could give would be greatly appreciated. Thank you very much. Tim Fallon Date: 11/17/97 at 16:44:28 From: Doctor Anthony Subject: Re: 3d geometry The general method is perhaps best illustrated by working through an example. Find the shortest distance between the given lines, and the points of closest approach on each line. x y-3 z x-5 y-8 z-2 --- = --- = --- = s and ----- = ----- = ------ = t 1 1 -1 3 7 -1 The common perpendicular is obtained from the vector product | i j k | = i(6) -j(2) + k(4) | 1 1 -1 | | 3 7 -1 | So common perpendicular is the vector (3, -1, 2) which can be written as a unit vector in the form 1/(sqrt(14)[3, -1, 2] The vector connecting the given point on line (1) with the given point on line (2) is [(5-0), (8-3), (2-0)] = (5, 5, 2) The scalar product of this with the common perpendicular in unit vector form is 5 x 3 + 5 x (-1) + 2 x 2 14 ------------------------- = ------- = sqrt(14) sqrt(14) sqrt(14) So the shortest distance is sqrt(14). Now to find the points of closest approach, we have: line (1) is r = (0, 3, 0) + s(1, 1, -1) = [s, (3+s), -s] line (2) is r = (5, 8, 2) + t(3, 7, -1) = [(5+3t), (8+7t), (2-t)] and we must find s and t. Line joining a general point on line(1) to a general point on line(2) is the vector [(5+3t-s), (8+7t-3-s), (2-t+s)] [(5+3t-s), (5+7t-s), (2-t+s)] and if s and t are points of closest approach this must be parallel to the vector (3, -1, 2) 5+3t-s 5+7t-s 2-t+s so ------ = -------- = ------ 3 -1 2 From these equations s = -1 and t = -1 The points of closest approach are therefore on line(1) (-1, 2, 1) and on line(2) (2, 1, 3) Check that shortest distance is sqrt(14) shortest distance = sqrt((2+1)^2 + (1-2)^2 + (3-1)^2) = sqrt(9 + 1 + 4) = sqrt(14) If you require the mid-point of these points, you get -1+2 2+1 1+3 ----, ----, ---- = [1/2, 3/2, 2] 2 2 2 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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