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.
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