Posted by **Mackenzie** on Tuesday, December 4, 2012 at 5:40pm.

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?

- Calculus -
**A.**, Tuesday, December 4, 2012 at 6:09pm
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.

- Calculus -
**A.**, Tuesday, December 4, 2012 at 6:12pm
*oops (5/2,1/2,0) for position

Looking at the answer again,

it looks like they divided the vector by 2

- Calculus -
**Mackenzie**, Tuesday, December 4, 2012 at 6:22pm
Do you know why they divided it by 2?

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