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Find the intersection point between two lines: L1=<-3,5,7>k+<1,-2,-0> L2=<3,-1,-4>t+<1,2,3>

Mathematics
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just equate the values of x, y, z from those two lines and find the valeus of k and t. if you don't find singular values ... then they do not intersect.
Hey @experimentX . I tried solving this intersection using a similar method you showed me for the line and plane intersection, does it work here as well?
Yeah, I found the intersection using the x,y,z components from L1 and L2 and found the correct intersection. I was just wondering if we could solve it using the other way you showed me for the line and plane problem

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yeah it works .... just try to find the value of k and t ... from first two equation. put the value on the last equation. If it's invalid then the line does not intersect.
Okay, so far I have from L2: 1+3t -->k1 2-t ---->k2 3-4t--->k3 and L1: -3k1+5k2+7k3=1 Filling those values from L2 into L1, I get: 1=-3(1+3t)+5(2-t)+7(3-4t) 1=-3-9t+10-5t+21-28t t=27/42
lol ... what are you doing?? |dw:1350576239788:dw|
Was what I did not the way i'm suppose to do it? lol
I just tried to do it a similar way to the way we solved the line and plane intersection; breaking up the line equation into it's x,y,z components, then filling them into the plane equation, except here there's two lines, so I broke up one of the line equations into it's x,y,z, and filled it into the other line equation
you are making it complicated... we are trying to find the point common to both lines. so just equate x, y and z of both lines.
Got it, thanks :D
well ... trying to find a plane that would contain both lines is not bad either. good pratice for other problems!

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