# Calculus

Find the parametric equations for the line of intersection of the planes x+y+z=3 and x-y+2z=2

I took the cross product of the 2 equations and got 3i-j-2k

I then set z=0 and got x=5/2 and y=1/2.
I got:
x=5/2 +3t
y=1/2-t
z=-2t

However, the answers are supposed to be:
x=5/2-(3/2)t
y=1/2+(1/2)t
z=t

What is my procedure missing to get there?

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1. I think your method is correct.
You took the cross product of the two vectors of the plane <1,1,1>x<1,-1,2> and got the direction vector v=<3,-1,-2>

You combined the two equations and got (0,5/2,1/2), which gives you the position

There are many different solutions for an equation of a line. So don't think that your answer is wrong.

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posted by A.
2. *oops (5/2,1/2,0) for position

it looks like they divided the vector by 2

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posted by A.
3. Do you know why they divided it by 2?

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