Shortest Distance between Points
Date: 01/17/98 at 17:58:53 From: Justin Subject: Shortest distance between lines I am doing a science fair project on the shortest distance between two points via another plane. I need help with my theorems. Thanks.
Date: 01/17/98 at 18:16:31 From: Doctor Anthony Subject: Re: Shortest distance between lines I'm not sure whether you really mean the shortest distance between two lines, as your subject line says, or between two points, as your question says. I'll address the question of the distance between two lines. I will illustrate the method by way of an example. I hope you can follow the working here. If not, write back. 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 the 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. The 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) Now, if your question is really that you want to find the distance between two lines while travelling along a given plane, simply find the two points where the two lines puncture the plane, then find the distance between those two points. -Doctors Anthony and Ken, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 08/125/98 at 2:11:26 From: Doctor Ujjwal Subject: Re: Shortest distance between lines In the subject line of your question you are asking about the distance between two lines, but in your actual question you ask about the shortest distance between two points. Here is an outline of how you would find the shortest distance between two POINTS via a plane. Let P = The given plane A, B = Two points whose distance via P is to be found. To find the distance 'via' P, we go from A to P and then to B. There are three cases: I. A and B are on opposite sides of P II. A or B (or both) lie in plane P III. A and B are on the same side of the plane | | | A* | A* | A* | \ | \ | \ | \ | \ | \ | \| \ | \| | \| *I |\ *B /| | \ | / | | *B | B* | | | | P P P CASE I CASE II CASE III CASES I & II are straightforward - the shortest distance from (A) to (B) via (P) is (AB). CASE III is very interesting. It is the classic case of reflection from a flat mirror! It is a well-known fact that light rays follow the shortest path during reflection. A ray from (A) reaches (B) after being reflected from mirror (P). So just find the 'reflection' (B') of (B) about (P) and then find the distance (AB') as in case I. | A* | \ | \ | \| *I /|\ / | \ B*--|--*B' | | O*--->N | | P CASE III The steps will be as follows: 0. The given formulation of the plane (P), may directly give (or help find) a point (O) in the plane and a normal (N) to the plane. 1. Normalize (N) to get the unit normal (n) 2. Find dot product (d) of vectors (n) and (OB). This is the signed distance of point (B) from plane (P). A positive sign means (B) is on the side of the plane pointed towards by the normal. 3. The 'image' or reflection (B') of point (B) about the 'mirror' or plane (P) can be found by walking twice the distance (d) 'into' the mirror from point (B). Thus, B' = B - 2(d)(n) .... substitute (d) with its proper sign. 4. Find the distance (AB') as usual. - Doctor Ujjwal Rane http://mathforum.org/dr.math/
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