Shortest Distance Between Two Vectors
Date: 04/07/98 at 02:39:55 From: blah Subject: Vectors How do you find the distance between the lines (1,2,3) + s(1,0,-1) and x = 0, y = 1+2t, z = 3+t? I think I have to change the lines to (1+s,2,3-s) and (0,1+2t,3+t) then do d = (|a1 - a2|.(b1 x b2))/|b1 x b2| where a1 and a2 are points on each of the lines, and b1 and b2 are the two lines. Is this right? Which two points should I use? Does it matter which points I pick? Thanks :) Mostyn
Date: 04/07/98 at 06:58:42 From: Doctor Anthony Subject: Re: Vectors To find the shortest distance between two skew lines (lines which do not meet) you first require the direction of the common perpendicular. Express this vector as a unit vector and then find the component in this direction of a line joining ANY two points on the two lines. This component will be the shortest distance between the lines. The direction of one line is: (1, 0, -1) and of the other: (0, 2, 1) You can use vector products to find the common perpendicular: |i j k| |1 0 -1| = i(2) - j(1) + k(2) = 2i - j + 2k |0 2 1| Express this as a unit vector: (1/3)[2i - j + 2k] (Equation 1) A line joining (1,2,3) on line 1 to (0,1,3) on line 2 is the vector: [(1-0), (2-1), (3-3)] = [1, 1, 0] Now scalar multiply this with the unit vector shown in Equation 1: Shortest distance = (1/3)[1*2 + 1*(-1) + 0*2] = (1/3)[2-1] = 1/3 units. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 04/07/98 at 09:07:37 From: blah Subject: Re: Vectors I can see that the direction of the lines is found by taking the coefficient of the parameter. Is there a formal way of doing this, or is it better just to learn the trick? Thanks again =) Mostyn
Date: 04/07/98 at 11:03:25 From: Doctor Anthony Subject: Re: Vectors The general vector equation of a line is: r = p + t*u where p is a point on the line and u is a vector in the direction of the line. To go from the origin to any point r1 on the line we go first to the point p, then a distance proportional to t along the vector u. So when the equation is written in the form (1,2,3) + s(1,0,-1), we know that (1,2,3) is a point on the line and (1,0,-1) is the direction of the line. If we write the direction vector as a unit vector this would be: r = (1,0,-1) + (s/sqrt(2))(1,0,-1) and now s is the actual distance along the line from the point p. If you see an equation of a line written in the form: x-x1 y-y1 z-z1 ---- = ---- = ----- = t a b c then (a,b,c) are direction ratios of the line, (x1,y1,z1) is a point on the line, and t is the same parameter as we used in the vector equation. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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