# vectors and the geometry of space

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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...

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I think the first three meet when (2t+1)=(3t+2)=(4t+3)

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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?

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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.

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