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Mathematics 18 Online
OpenStudy (anonymous):

For x+y+z=1 and x+2y+2z=1, find the parametric equations for the line of intersection of the planes

OpenStudy (anonymous):

I know that I'd have to do the cross product of those two planes, which I'm perfectly fine with that part. It's a matter of finding the point after setting one variable to zero that I'm a bit rusty on. Professor kinda rushed it in class and skipped this step

OpenStudy (loser66):

Show your work, please

OpenStudy (anonymous):

x+y+z=1 gives n1=<1,1,1> x+2y+2z=1 gives n2=<1,2,2> Then I would take the cross product: n=n1xn2 Following this, whatever point I get from the issue mentioned above by setting one variable=0 and solving for the other two to get some point (x,y,z) would be multiplied by the vector n - then it's just a matter of formatting. I just don't really remember the process to get the point and we did this right at the very end of class (he went over the time), so the process was not discussed in a way that would refresh my memory. I understand the rest just fine

OpenStudy (loser66):

ok, I give you this x +y + z = 1 x + 2y +2z = 1 hence y + z =0 or y = -z. set z = t, then y = -t x = 1 -y - z = 1+t +t = 1+ 2t Your parametric equation is x = 1 + 2t y = -t z =t dat sit

OpenStudy (anonymous):

That's all I needed right there, got it now - thank you!

OpenStudy (loser66):

To this kind of problem, just add them up or subtract them to eliminate one variable, then set z = t to solve for y, then for x. Dat sit

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