Find the vector equation of the line of intersection for the pair of planes.

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'

1 answer

You 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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