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July 29, 2014

July 29, 2014

Posted by **Shaila** on Wednesday, September 1, 2010 at 3:08am.

Plane one: x+5y-3z-8=0

Plane two: y+2z-4=0

I did half of the work but now i am stuck.

the normal of the planes are not parallel and therefore a solution exists, you simultaneously solve the equations. I got -13z+12=0. Now if i let z = t then how do i find the 'y' and the 'x'

- Math - Intersection of planes -
**MathMate**, Wednesday, September 1, 2010 at 7:52amYou are almost there.

If you look at the system of equations of the planes

Plane one: x+5y-3z-8=0

Plane two: y+2z-4=0

when converted to the reduced echelon form, you get

x + 0y -13z = -12

0x + y + 2z = 4

Take z as the free variable, (i.e. t) and solve for x and y in terms of t, and z equals t naturally. You will get the solution vector of (x,y,z) in terms of t. Transform this to the appropriate form for your answer.

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