caseclosed Posted September 10, 2006 Share Posted September 10, 2006 find the point of intersection of the lines x=2t+1, y=3t+2, z=4t+3, and x=s+2, y=2s+4, z=-4s-1, and the find the plane determined by these lines. I don't know what to do... Link to comment Share on other sites More sharing options...
the tree Posted September 10, 2006 Share Posted September 10, 2006 I think the first three meet when (2t+1)=(3t+2)=(4t+3) Link to comment Share on other sites More sharing options...
matt grime Posted September 10, 2006 Share Posted September 10, 2006 That can't be correct. Why would they meet at a point when x=y=z necessarily? And you can solve that for t and get t=1 but that doesn't have anything to do with s and the second line, does it? You just need to find a point, if any exists, that lies on both lines, so do it. When are the x coordinates the same? And the y and z coordinates. Those are conditions on s and t. Now can they be simultaneously satisfied? Link to comment Share on other sites More sharing options...
the tree Posted September 10, 2006 Share Posted September 10, 2006 Aha, that is what happens when I'm caught between looking at maths and going to lunch. Obviously the intersection is where (2t+1)=(s+2) etc. Apologies for my carelessness resulting in a pretty useless post. Link to comment Share on other sites More sharing options...
ajb Posted September 10, 2006 Share Posted September 10, 2006 This may help http://mathworld.wolfram.com/Line-LineIntersection.html http://mathworld.wolfram.com/Line-PlaneIntersection.html Link to comment Share on other sites More sharing options...
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