Intersection Point of Two LinesDate: 07/22/2003 at 10:19:15 From: Bensegueni Subject: How to find the intersection point of two lines in 3D I want to find the intersection point of two lines (in 3D) defined by their direction vectors V1 and V2. I found this question in the archive: Vector Algebra: Finding the Intersection Point http://mathforum.org/library/drmath/view/62814.html but when I used the method given it didn't work. The method was to use the two equations that represent the lines: L1 = P1 + aV1 L2 = P2 + bV2 Where V1 and V2 are the director vectors P1 point which belongs to L1 P2 point which belongs to L2 <=> P1 + aV1 = P2 + bV2 => aV1 = (P2-P1) + bV2 x V2 in the two sides we will have (cross product): a(V1 x V2) = (P2-P1) x V2 once we have "a" we can replace it in the 1st equation to get the point of intersection, but I don't get a correct result. Date: 07/22/2003 at 13:00:14 From: Doctor George Subject: Re: how to find the intersection point of two lines in 3D Hi Bensegueni, Here is another way to think about intersecting two lines in 3D. Define line 1 to contain point (x1,y1,z1) with vector (a1,b1,c1). Define line 2 to contain point (x2,y2,z2) with vector (a2,b2,c2). We can write these parametric equations for the lines. Line1 Line2 ----- ----- x = x1 + a1 * t1 x = x2 + a2 * t2 y = y1 + b1 * t1 y = y2 + b2 * t2 z = z1 + c1 * t1 z = z2 + c2 * t2 If we set the two x values equal, and the two y values equal we get these two equations. x1 + a1 * t1 = x2 + a2 * t2 y1 + b1 * t1 = y2 + b2 * t2 You can solve these equations for t1 and t2. Then put those values back into the parametric equations to solve for the intersection point. If you have done the arithmetic correctly, you only need to use one of the equations for each of x and y. You should check both equations for z to make sure they give the same result. If they give different results then the lines are skew. Does that make sense? Write again if you need more help. - Doctor George, The Math Forum http://mathforum.org/dr.math/ Date: 02/13/2008 at 13:59:39 From: Ean Subject: A fallacious answer? I've tried the method for finding the intersection of two lines in 3D that was given in your archive answer 63719 but it doesn't seem to work. Can you show me an example of how it works? Date: 02/13/2008 at 15:44:04 From: Doctor Peterson Subject: Re: A fallacious answer? Solving for two unlike terms... Hi, Ean. Thanks for writing to Dr. Math. Here's an example; I'll use both the method the questioner showed (which Dr. George never commented on) and Dr. George's method, and maybe you can tell me where you get something erroneous. Take P1 = (1,0,0) V1 = (2,3,1) P2 = (0,5,5) V2 = (5,1,-3) Using Bensegueni's method, we find a(V1 x V2) = a(-10,-11,-13) and (P2-P1) x V2 = (-1,5,5)x(5,1,-3) = (-20,-22,-26) Since the latter is 2(-10,-11,-13), we have a = 2, and the point of intersection is P1 + 2V1 = (1,0,0)+(4,6,2) = (5,6,2) To find b and check the answer, P2 + bV2 = (0,5,5) + b(5,1,-3) which will equal (5,6,2) when b=1. So we have a solution. Now try Dr. George's method. The lines are Line1 Line2 ----- ----- x = 1 + 2t1 x = 0 + 5t2 y = 0 + 3t1 y = 5 + 1t2 z = 0 + 1t1 z = 5 - 3t2 Setting the x's and y's equal, we have to solve 1 + 2t1 = 0 + 5t2 0 + 3t1 = 5 + 1t2 This reduces to 2t1 - 5t2 = -1 3t1 - 1t2 = 5 We can multiply the first equation by -1 and the second equation by 5 and add: -2t1 + 5t2 = 1 15t1 - 5t2 = 25 --------------- 13t1 = 26 t1 = 2 Plugging that into the second of our pair, we get 3*2 - t2 = 5 which gives t2 = 1 Plugging those into all six equations for x, y, and z, we get x = 1 + 2*2 = 5 x = 0 + 5*1 = 5 y = 0 + 3*2 = 6 y = 5 + 1*1 = 6 z = 0 + 1*2 = 2 z = 5 - 3*1 = 2 So this is indeed the intersection of the lines. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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