Distance between LinesDate: 03/04/97 at 23:22:55 From: Brandon Gee Subject: Geometrical applications with vectors Dear Dr. Math, I am in analysis and am struggling. I need help with just problem 6b. I wrote the other questions because 6b refers to some of them, and I was not sure which ones, so I wrote them all. 1. A Cartesian coordinate system is set up with its origin in the corner, its x- and y- axes along the intersection of the floor with each wall, and its z- axis vertical, along the intersection of the two walls. Its x-, y-, and z-intercepts are 12, 20, and 9 m, respectively. (a) Sketch the portion of the plane that lies in the first octant. (b) Find the vector normal to the plane in part (a) with the smallest possible integer coeficients. (c) Find an equation for the plane in part (a). (d) What is the closest the plane in part (a) comes to the origin? 2. Three vertical columns are to be built from the floor at points (x,y) = (1,5), (2,10) and (5,5), going up to the plane in Problem 1. (a) Find the length of each column (b) The three columns pierce the plane at points P(sub 1), P(sub 2), and P(sub 3) respectively. Sketch the columns and the triangle with these points as vertices in your figure from Problem 1. (c) How many square meters of paint would be needed to paint the triangle in part (b)? 5. A wire is to be run along the line containing point P(sub 1) and P(sub 2) from Problem 2. (a) Find the vector equation of this line. (b) Where does the wire pierce the floor? 6. Another wire runs along the line with a vector equation: r = (5 + (2/5)d)i - (7 - (2/3)d)j + (2 - (1/3)d)k (a) Find the point where this wire pierces the plane in Problem 1. (b) How close does this wire come to the wire in problem 5? I have no clue how to do this, or even where to start. I would greatly appreciate it if you could answer this question in its entirety along with a brief explanation of each step. If I am asking too much, the answer will suffice. Thank you, Brandon Gee Date: 03/05/97 at 08:01:47 From: Doctor Jerry Subject: Re: Geometrical applications with vectors Hi Brandon, I'll try to describe how to find the distance between two lines. I'd like to write them a little differently, as r = a+t*b, where a is a point through which the line passes and b is a vector in the direction of the line. For example, r={2,1,3} + t*{1,1,-1} is a line through the point with position vector {2,1,3} and in the direction of the vector {1,1,-1}. I think you would write this as: r = (2+t)i + (1+t)j + (3-t)k. line 1: r = a + t*b line 2: r = p +s*q The parameters for the lines are t and s, though they won't be used in the method I'll describe. Imagine, for a moment, that these two lines have been moved so that they intersect. During the moves, their directions do not change (they are moving "parallel to themselves"). They then define a plane. A vector perpendicular or normal to this plane is b x q (b cross q). The distance between line 1 and line 2 is measured along the direction b x q. Let u = (1/|b x q|) b x q, where |b x q| is the length of b x q. Now, u is a unit vector. Take any point m on line 1 and any point n on line 2. The distance between the lines is |(m-n).u| (I'm using . as dot product). For m and n you can take m = a and n = p. On line 1, if t = 0, r = a, so that a is a point on line 1. Similarly for line 2. Why does this formula work? Imagine a thin cord connecting m and n. What we have done is to calculate the vector m-n, which stretches from one end of the cord to the other end. The ends of the cord lie on the two lines. Then we project m-n onto the unit vector u, which is perpendicular to both lines. If line 1 is {2,1,3} + t{1,1,-1} and line 2 is {8,9,0} + s{2,3,2}, then u = (1/sqrt(42)){-5,4,-1} and the distance between these lines is: |({2,1,3}-{8,9,0}).u| = 5/sqrt(42) -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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